149651
A hot container takes $1 \mathrm{~min}$ to cool from $95^{\circ} \mathrm{C}$ to $75^{\circ} \mathrm{C}$. The time it takes to cool from $74^{\circ} \mathrm{C}$ to $54^{\circ} \mathrm{C}$ is [Consider the room temperature as $30^{\circ} \mathrm{C}$ ]
1 $97 \mathrm{~s}$
2 $70 \mathrm{~s}$
3 $102 \mathrm{~s}$
4 $82 \mathrm{~s}$
Explanation:
A Given, initial Temperature $\left(\theta_{1}\right)=95^{\circ} \mathrm{C}$ Final Temperature $\left(\theta_{2}\right)=75^{\circ} \mathrm{C}$ Atmospheric Temperature $(\theta)=30^{\circ} \mathrm{C}$ Time $(\mathrm{t})=1 \mathrm{~min}$ According to Newton's cooling laws $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=-\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta\right)$ $\frac{95-75}{1}=-\mathrm{K}\left(\frac{95+75}{2}-30\right)$ $20=-\mathrm{K}(85-30)$ $\mathrm{K}=-\frac{20}{55}$ Second case, $\theta_{3}=74^{\circ} \mathrm{C}, \theta_{4}=54^{\circ} \mathrm{C}, \theta=30^{\circ} \mathrm{C}$ $\mathrm{K}=-(20 / 55)$ Again use Newton's cooling law, $\frac{74-54}{\mathrm{t}}=-\left(-\frac{20}{55}\right)\left(\frac{74+54}{2}-30\right)$ $\frac{20}{\mathrm{t}}=\frac{20}{55}(64-30)$ $\frac{20}{\mathrm{t}}=\frac{20}{55} \times 34 \Rightarrow \mathrm{t}=\frac{55}{34} \mathrm{~min}$ $=\frac{55}{34} \times 60=97.06 \mathrm{sec} .$
TS EAMCET 30.07.2022
Heat Transfer
149652
A body cools from $60{ }^{\circ} \mathrm{C}$ to $40{ }^{\circ} \mathrm{C}$ in the first 7 minutes and to $28^{\circ} \mathrm{C}$ in the next 7 minutes. The temperature of the surroundings is
1 $10^{\circ} \mathrm{C}$
2 $20^{\circ} \mathrm{C}$
3 $5{ }^{\circ} \mathrm{C}$
4 $30^{\circ} \mathrm{C}$
Explanation:
A From Newton's Law of cooling, $\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{t}}=-\mathrm{K}\left[\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}\right)-\mathrm{T}_{\mathrm{o}}\right]$ $\frac{60-40}{7 \times 60}=-\mathrm{K}\left[\left(\frac{60+40}{2}-\mathrm{T}_{\mathrm{o}}\right)\right.$ $\frac{20}{420}=-\mathrm{K}\left(50-\mathrm{T}_{\mathrm{o}}\right)$ Again cool 7 min, $\frac{40-28}{7 \times 60}=-\mathrm{K}\left[\left(\frac{40+28}{2}\right)-\mathrm{T}_{\mathrm{o}}\right]$ $\frac{12}{420}=-\mathrm{K}\left(34-\mathrm{T}_{\mathrm{o}}\right)$ Dividing Equation (i) by (ii), we get - $\frac{20}{12}=\frac{50-\mathrm{T}_{\mathrm{o}}}{34-\mathrm{T}_{\mathrm{o}}}$ $680-20 \mathrm{~T}_{\mathrm{o}}=600-12 \mathrm{~T}_{\mathrm{o}}$ $80=8 \mathrm{~T}_{\mathrm{o}}$ $\mathrm{T}_{\mathrm{o}}=10^{\circ} \mathrm{C}$
AP EAMCET-24.04.2019
Heat Transfer
149653
One end of a metal rod of length $1.0 \mathrm{~m}$ and area of cross-section $1 \mathrm{~m}^{2}$ is maintained at $100^{\circ} \mathrm{C}$ : If the other end of the rod is maintained at $0^{\circ} \mathrm{C}$, the quantity of heat transmitted through the rod per minute is:
1 $3 \times 10^{3} \mathrm{~J}$
2 $6 \times 10^{3} \mathrm{~J}$
3 $9 \times 10^{3} \mathrm{~J}$
4 $12 \times 10^{3} \mathrm{~J}$
Explanation:
B Given that, Length of $\operatorname{Rod}=1.0 \mathrm{~m}$ Area of cross-section $=1 \mathrm{~m}^{2}$ Temperature $(\Delta t)=100^{\circ} \mathrm{C}$ As we know that, Heat transmitted through the Rod, $\frac{\mathrm{Q}}{\mathrm{t}}=\frac{\mathrm{kA} \Delta \mathrm{t}}{\mathrm{l}}$ $=\frac{1 \times 1 \times(100-0)}{1}$ $\frac{\mathrm{Q}}{60}=100$ $\mathrm{Q}=60 \times 100$ $=6 \times 10^{3}$ Joule
AP EAMCET(Medical)-2000
Heat Transfer
149654
A copper sphere cools from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 10 minutes and to $42^{\circ} \mathrm{C}$ in the next 10 minutes. Calculate the temperature of the surrounding.
1 $18.01^{\circ} \mathrm{C}$
2 $26^{\circ} \mathrm{C}$
3 $10.6^{\circ} \mathrm{C}$
4 $20^{\circ} \mathrm{C}$
Explanation:
B From the Newton's law of cooling. $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right]$ Given, $62^{\circ} \mathrm{C}-50^{\circ} \mathrm{C} \text { in } 10 \mathrm{~min}$ $50^{\circ} \mathrm{C}-42^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$ Then Temperature of surrounding $\theta_{0}=$ ? $\frac{62-50}{10}=K\left[\frac{62+50}{2}-\theta_{0}\right]$ $\frac{50-42}{10}=K\left[\frac{50+42}{2}-\theta_{0}\right]$ Equation (i) / (ii) $\frac{12}{8}=\frac{56-\theta_{0}}{46-\theta_{0}}$ $12 \times 46-12 \theta_{0}=56 \times 8-8 \theta_{0}$ $552-12 \theta_{0}=448-8 \theta_{0}$ $4 \theta_{0}=552-448$ $4 \theta_{0}=104$ $\theta_{0}=\frac{104}{4}$ $\theta_{0}=26^{\circ} \mathrm{C}$
BITSAT-2011
Heat Transfer
149655
A body cools in $7 \mathrm{~min}$ from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. What times (in min) does it take to cool from $40^{\circ} \mathrm{C}$ to $28^{\circ} \mathrm{C}$, if surrounding temperature is $10^{\circ} \mathrm{C}$ ? (Assume Newton's law of cooling)
1 3.5
2 14
3 7
4 10
Explanation:
C Given, $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{Cin} 7 \mathrm{~min}$ $40^{\circ} \mathrm{C}$ to $28^{\circ} \mathrm{C}$ in $\mathrm{t}$ min Given $\mathrm{T}_{\text {surrounding }}=10^{\circ} \mathrm{C}$ Newton's law of cooling $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right)$ Putting the value in equation (i) $\frac{60-40}{7}=\mathrm{K}\left(\frac{60+40}{2}-10\right)$ $\frac{40-28}{\mathrm{t}}=\mathrm{K}\left(\frac{40+28}{2}-10\right)$ Dividing eq $^{\mathrm{n}}$ (ii) / (iii) $\frac{t \times 20}{7 \times 12}=\frac{40}{24}$ $\frac{5 t}{7 \times 3}=\frac{5}{3}$ $t=7 \text { min. }$
149651
A hot container takes $1 \mathrm{~min}$ to cool from $95^{\circ} \mathrm{C}$ to $75^{\circ} \mathrm{C}$. The time it takes to cool from $74^{\circ} \mathrm{C}$ to $54^{\circ} \mathrm{C}$ is [Consider the room temperature as $30^{\circ} \mathrm{C}$ ]
1 $97 \mathrm{~s}$
2 $70 \mathrm{~s}$
3 $102 \mathrm{~s}$
4 $82 \mathrm{~s}$
Explanation:
A Given, initial Temperature $\left(\theta_{1}\right)=95^{\circ} \mathrm{C}$ Final Temperature $\left(\theta_{2}\right)=75^{\circ} \mathrm{C}$ Atmospheric Temperature $(\theta)=30^{\circ} \mathrm{C}$ Time $(\mathrm{t})=1 \mathrm{~min}$ According to Newton's cooling laws $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=-\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta\right)$ $\frac{95-75}{1}=-\mathrm{K}\left(\frac{95+75}{2}-30\right)$ $20=-\mathrm{K}(85-30)$ $\mathrm{K}=-\frac{20}{55}$ Second case, $\theta_{3}=74^{\circ} \mathrm{C}, \theta_{4}=54^{\circ} \mathrm{C}, \theta=30^{\circ} \mathrm{C}$ $\mathrm{K}=-(20 / 55)$ Again use Newton's cooling law, $\frac{74-54}{\mathrm{t}}=-\left(-\frac{20}{55}\right)\left(\frac{74+54}{2}-30\right)$ $\frac{20}{\mathrm{t}}=\frac{20}{55}(64-30)$ $\frac{20}{\mathrm{t}}=\frac{20}{55} \times 34 \Rightarrow \mathrm{t}=\frac{55}{34} \mathrm{~min}$ $=\frac{55}{34} \times 60=97.06 \mathrm{sec} .$
TS EAMCET 30.07.2022
Heat Transfer
149652
A body cools from $60{ }^{\circ} \mathrm{C}$ to $40{ }^{\circ} \mathrm{C}$ in the first 7 minutes and to $28^{\circ} \mathrm{C}$ in the next 7 minutes. The temperature of the surroundings is
1 $10^{\circ} \mathrm{C}$
2 $20^{\circ} \mathrm{C}$
3 $5{ }^{\circ} \mathrm{C}$
4 $30^{\circ} \mathrm{C}$
Explanation:
A From Newton's Law of cooling, $\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{t}}=-\mathrm{K}\left[\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}\right)-\mathrm{T}_{\mathrm{o}}\right]$ $\frac{60-40}{7 \times 60}=-\mathrm{K}\left[\left(\frac{60+40}{2}-\mathrm{T}_{\mathrm{o}}\right)\right.$ $\frac{20}{420}=-\mathrm{K}\left(50-\mathrm{T}_{\mathrm{o}}\right)$ Again cool 7 min, $\frac{40-28}{7 \times 60}=-\mathrm{K}\left[\left(\frac{40+28}{2}\right)-\mathrm{T}_{\mathrm{o}}\right]$ $\frac{12}{420}=-\mathrm{K}\left(34-\mathrm{T}_{\mathrm{o}}\right)$ Dividing Equation (i) by (ii), we get - $\frac{20}{12}=\frac{50-\mathrm{T}_{\mathrm{o}}}{34-\mathrm{T}_{\mathrm{o}}}$ $680-20 \mathrm{~T}_{\mathrm{o}}=600-12 \mathrm{~T}_{\mathrm{o}}$ $80=8 \mathrm{~T}_{\mathrm{o}}$ $\mathrm{T}_{\mathrm{o}}=10^{\circ} \mathrm{C}$
AP EAMCET-24.04.2019
Heat Transfer
149653
One end of a metal rod of length $1.0 \mathrm{~m}$ and area of cross-section $1 \mathrm{~m}^{2}$ is maintained at $100^{\circ} \mathrm{C}$ : If the other end of the rod is maintained at $0^{\circ} \mathrm{C}$, the quantity of heat transmitted through the rod per minute is:
1 $3 \times 10^{3} \mathrm{~J}$
2 $6 \times 10^{3} \mathrm{~J}$
3 $9 \times 10^{3} \mathrm{~J}$
4 $12 \times 10^{3} \mathrm{~J}$
Explanation:
B Given that, Length of $\operatorname{Rod}=1.0 \mathrm{~m}$ Area of cross-section $=1 \mathrm{~m}^{2}$ Temperature $(\Delta t)=100^{\circ} \mathrm{C}$ As we know that, Heat transmitted through the Rod, $\frac{\mathrm{Q}}{\mathrm{t}}=\frac{\mathrm{kA} \Delta \mathrm{t}}{\mathrm{l}}$ $=\frac{1 \times 1 \times(100-0)}{1}$ $\frac{\mathrm{Q}}{60}=100$ $\mathrm{Q}=60 \times 100$ $=6 \times 10^{3}$ Joule
AP EAMCET(Medical)-2000
Heat Transfer
149654
A copper sphere cools from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 10 minutes and to $42^{\circ} \mathrm{C}$ in the next 10 minutes. Calculate the temperature of the surrounding.
1 $18.01^{\circ} \mathrm{C}$
2 $26^{\circ} \mathrm{C}$
3 $10.6^{\circ} \mathrm{C}$
4 $20^{\circ} \mathrm{C}$
Explanation:
B From the Newton's law of cooling. $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right]$ Given, $62^{\circ} \mathrm{C}-50^{\circ} \mathrm{C} \text { in } 10 \mathrm{~min}$ $50^{\circ} \mathrm{C}-42^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$ Then Temperature of surrounding $\theta_{0}=$ ? $\frac{62-50}{10}=K\left[\frac{62+50}{2}-\theta_{0}\right]$ $\frac{50-42}{10}=K\left[\frac{50+42}{2}-\theta_{0}\right]$ Equation (i) / (ii) $\frac{12}{8}=\frac{56-\theta_{0}}{46-\theta_{0}}$ $12 \times 46-12 \theta_{0}=56 \times 8-8 \theta_{0}$ $552-12 \theta_{0}=448-8 \theta_{0}$ $4 \theta_{0}=552-448$ $4 \theta_{0}=104$ $\theta_{0}=\frac{104}{4}$ $\theta_{0}=26^{\circ} \mathrm{C}$
BITSAT-2011
Heat Transfer
149655
A body cools in $7 \mathrm{~min}$ from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. What times (in min) does it take to cool from $40^{\circ} \mathrm{C}$ to $28^{\circ} \mathrm{C}$, if surrounding temperature is $10^{\circ} \mathrm{C}$ ? (Assume Newton's law of cooling)
1 3.5
2 14
3 7
4 10
Explanation:
C Given, $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{Cin} 7 \mathrm{~min}$ $40^{\circ} \mathrm{C}$ to $28^{\circ} \mathrm{C}$ in $\mathrm{t}$ min Given $\mathrm{T}_{\text {surrounding }}=10^{\circ} \mathrm{C}$ Newton's law of cooling $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right)$ Putting the value in equation (i) $\frac{60-40}{7}=\mathrm{K}\left(\frac{60+40}{2}-10\right)$ $\frac{40-28}{\mathrm{t}}=\mathrm{K}\left(\frac{40+28}{2}-10\right)$ Dividing eq $^{\mathrm{n}}$ (ii) / (iii) $\frac{t \times 20}{7 \times 12}=\frac{40}{24}$ $\frac{5 t}{7 \times 3}=\frac{5}{3}$ $t=7 \text { min. }$
149651
A hot container takes $1 \mathrm{~min}$ to cool from $95^{\circ} \mathrm{C}$ to $75^{\circ} \mathrm{C}$. The time it takes to cool from $74^{\circ} \mathrm{C}$ to $54^{\circ} \mathrm{C}$ is [Consider the room temperature as $30^{\circ} \mathrm{C}$ ]
1 $97 \mathrm{~s}$
2 $70 \mathrm{~s}$
3 $102 \mathrm{~s}$
4 $82 \mathrm{~s}$
Explanation:
A Given, initial Temperature $\left(\theta_{1}\right)=95^{\circ} \mathrm{C}$ Final Temperature $\left(\theta_{2}\right)=75^{\circ} \mathrm{C}$ Atmospheric Temperature $(\theta)=30^{\circ} \mathrm{C}$ Time $(\mathrm{t})=1 \mathrm{~min}$ According to Newton's cooling laws $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=-\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta\right)$ $\frac{95-75}{1}=-\mathrm{K}\left(\frac{95+75}{2}-30\right)$ $20=-\mathrm{K}(85-30)$ $\mathrm{K}=-\frac{20}{55}$ Second case, $\theta_{3}=74^{\circ} \mathrm{C}, \theta_{4}=54^{\circ} \mathrm{C}, \theta=30^{\circ} \mathrm{C}$ $\mathrm{K}=-(20 / 55)$ Again use Newton's cooling law, $\frac{74-54}{\mathrm{t}}=-\left(-\frac{20}{55}\right)\left(\frac{74+54}{2}-30\right)$ $\frac{20}{\mathrm{t}}=\frac{20}{55}(64-30)$ $\frac{20}{\mathrm{t}}=\frac{20}{55} \times 34 \Rightarrow \mathrm{t}=\frac{55}{34} \mathrm{~min}$ $=\frac{55}{34} \times 60=97.06 \mathrm{sec} .$
TS EAMCET 30.07.2022
Heat Transfer
149652
A body cools from $60{ }^{\circ} \mathrm{C}$ to $40{ }^{\circ} \mathrm{C}$ in the first 7 minutes and to $28^{\circ} \mathrm{C}$ in the next 7 minutes. The temperature of the surroundings is
1 $10^{\circ} \mathrm{C}$
2 $20^{\circ} \mathrm{C}$
3 $5{ }^{\circ} \mathrm{C}$
4 $30^{\circ} \mathrm{C}$
Explanation:
A From Newton's Law of cooling, $\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{t}}=-\mathrm{K}\left[\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}\right)-\mathrm{T}_{\mathrm{o}}\right]$ $\frac{60-40}{7 \times 60}=-\mathrm{K}\left[\left(\frac{60+40}{2}-\mathrm{T}_{\mathrm{o}}\right)\right.$ $\frac{20}{420}=-\mathrm{K}\left(50-\mathrm{T}_{\mathrm{o}}\right)$ Again cool 7 min, $\frac{40-28}{7 \times 60}=-\mathrm{K}\left[\left(\frac{40+28}{2}\right)-\mathrm{T}_{\mathrm{o}}\right]$ $\frac{12}{420}=-\mathrm{K}\left(34-\mathrm{T}_{\mathrm{o}}\right)$ Dividing Equation (i) by (ii), we get - $\frac{20}{12}=\frac{50-\mathrm{T}_{\mathrm{o}}}{34-\mathrm{T}_{\mathrm{o}}}$ $680-20 \mathrm{~T}_{\mathrm{o}}=600-12 \mathrm{~T}_{\mathrm{o}}$ $80=8 \mathrm{~T}_{\mathrm{o}}$ $\mathrm{T}_{\mathrm{o}}=10^{\circ} \mathrm{C}$
AP EAMCET-24.04.2019
Heat Transfer
149653
One end of a metal rod of length $1.0 \mathrm{~m}$ and area of cross-section $1 \mathrm{~m}^{2}$ is maintained at $100^{\circ} \mathrm{C}$ : If the other end of the rod is maintained at $0^{\circ} \mathrm{C}$, the quantity of heat transmitted through the rod per minute is:
1 $3 \times 10^{3} \mathrm{~J}$
2 $6 \times 10^{3} \mathrm{~J}$
3 $9 \times 10^{3} \mathrm{~J}$
4 $12 \times 10^{3} \mathrm{~J}$
Explanation:
B Given that, Length of $\operatorname{Rod}=1.0 \mathrm{~m}$ Area of cross-section $=1 \mathrm{~m}^{2}$ Temperature $(\Delta t)=100^{\circ} \mathrm{C}$ As we know that, Heat transmitted through the Rod, $\frac{\mathrm{Q}}{\mathrm{t}}=\frac{\mathrm{kA} \Delta \mathrm{t}}{\mathrm{l}}$ $=\frac{1 \times 1 \times(100-0)}{1}$ $\frac{\mathrm{Q}}{60}=100$ $\mathrm{Q}=60 \times 100$ $=6 \times 10^{3}$ Joule
AP EAMCET(Medical)-2000
Heat Transfer
149654
A copper sphere cools from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 10 minutes and to $42^{\circ} \mathrm{C}$ in the next 10 minutes. Calculate the temperature of the surrounding.
1 $18.01^{\circ} \mathrm{C}$
2 $26^{\circ} \mathrm{C}$
3 $10.6^{\circ} \mathrm{C}$
4 $20^{\circ} \mathrm{C}$
Explanation:
B From the Newton's law of cooling. $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right]$ Given, $62^{\circ} \mathrm{C}-50^{\circ} \mathrm{C} \text { in } 10 \mathrm{~min}$ $50^{\circ} \mathrm{C}-42^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$ Then Temperature of surrounding $\theta_{0}=$ ? $\frac{62-50}{10}=K\left[\frac{62+50}{2}-\theta_{0}\right]$ $\frac{50-42}{10}=K\left[\frac{50+42}{2}-\theta_{0}\right]$ Equation (i) / (ii) $\frac{12}{8}=\frac{56-\theta_{0}}{46-\theta_{0}}$ $12 \times 46-12 \theta_{0}=56 \times 8-8 \theta_{0}$ $552-12 \theta_{0}=448-8 \theta_{0}$ $4 \theta_{0}=552-448$ $4 \theta_{0}=104$ $\theta_{0}=\frac{104}{4}$ $\theta_{0}=26^{\circ} \mathrm{C}$
BITSAT-2011
Heat Transfer
149655
A body cools in $7 \mathrm{~min}$ from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. What times (in min) does it take to cool from $40^{\circ} \mathrm{C}$ to $28^{\circ} \mathrm{C}$, if surrounding temperature is $10^{\circ} \mathrm{C}$ ? (Assume Newton's law of cooling)
1 3.5
2 14
3 7
4 10
Explanation:
C Given, $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{Cin} 7 \mathrm{~min}$ $40^{\circ} \mathrm{C}$ to $28^{\circ} \mathrm{C}$ in $\mathrm{t}$ min Given $\mathrm{T}_{\text {surrounding }}=10^{\circ} \mathrm{C}$ Newton's law of cooling $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right)$ Putting the value in equation (i) $\frac{60-40}{7}=\mathrm{K}\left(\frac{60+40}{2}-10\right)$ $\frac{40-28}{\mathrm{t}}=\mathrm{K}\left(\frac{40+28}{2}-10\right)$ Dividing eq $^{\mathrm{n}}$ (ii) / (iii) $\frac{t \times 20}{7 \times 12}=\frac{40}{24}$ $\frac{5 t}{7 \times 3}=\frac{5}{3}$ $t=7 \text { min. }$
149651
A hot container takes $1 \mathrm{~min}$ to cool from $95^{\circ} \mathrm{C}$ to $75^{\circ} \mathrm{C}$. The time it takes to cool from $74^{\circ} \mathrm{C}$ to $54^{\circ} \mathrm{C}$ is [Consider the room temperature as $30^{\circ} \mathrm{C}$ ]
1 $97 \mathrm{~s}$
2 $70 \mathrm{~s}$
3 $102 \mathrm{~s}$
4 $82 \mathrm{~s}$
Explanation:
A Given, initial Temperature $\left(\theta_{1}\right)=95^{\circ} \mathrm{C}$ Final Temperature $\left(\theta_{2}\right)=75^{\circ} \mathrm{C}$ Atmospheric Temperature $(\theta)=30^{\circ} \mathrm{C}$ Time $(\mathrm{t})=1 \mathrm{~min}$ According to Newton's cooling laws $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=-\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta\right)$ $\frac{95-75}{1}=-\mathrm{K}\left(\frac{95+75}{2}-30\right)$ $20=-\mathrm{K}(85-30)$ $\mathrm{K}=-\frac{20}{55}$ Second case, $\theta_{3}=74^{\circ} \mathrm{C}, \theta_{4}=54^{\circ} \mathrm{C}, \theta=30^{\circ} \mathrm{C}$ $\mathrm{K}=-(20 / 55)$ Again use Newton's cooling law, $\frac{74-54}{\mathrm{t}}=-\left(-\frac{20}{55}\right)\left(\frac{74+54}{2}-30\right)$ $\frac{20}{\mathrm{t}}=\frac{20}{55}(64-30)$ $\frac{20}{\mathrm{t}}=\frac{20}{55} \times 34 \Rightarrow \mathrm{t}=\frac{55}{34} \mathrm{~min}$ $=\frac{55}{34} \times 60=97.06 \mathrm{sec} .$
TS EAMCET 30.07.2022
Heat Transfer
149652
A body cools from $60{ }^{\circ} \mathrm{C}$ to $40{ }^{\circ} \mathrm{C}$ in the first 7 minutes and to $28^{\circ} \mathrm{C}$ in the next 7 minutes. The temperature of the surroundings is
1 $10^{\circ} \mathrm{C}$
2 $20^{\circ} \mathrm{C}$
3 $5{ }^{\circ} \mathrm{C}$
4 $30^{\circ} \mathrm{C}$
Explanation:
A From Newton's Law of cooling, $\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{t}}=-\mathrm{K}\left[\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}\right)-\mathrm{T}_{\mathrm{o}}\right]$ $\frac{60-40}{7 \times 60}=-\mathrm{K}\left[\left(\frac{60+40}{2}-\mathrm{T}_{\mathrm{o}}\right)\right.$ $\frac{20}{420}=-\mathrm{K}\left(50-\mathrm{T}_{\mathrm{o}}\right)$ Again cool 7 min, $\frac{40-28}{7 \times 60}=-\mathrm{K}\left[\left(\frac{40+28}{2}\right)-\mathrm{T}_{\mathrm{o}}\right]$ $\frac{12}{420}=-\mathrm{K}\left(34-\mathrm{T}_{\mathrm{o}}\right)$ Dividing Equation (i) by (ii), we get - $\frac{20}{12}=\frac{50-\mathrm{T}_{\mathrm{o}}}{34-\mathrm{T}_{\mathrm{o}}}$ $680-20 \mathrm{~T}_{\mathrm{o}}=600-12 \mathrm{~T}_{\mathrm{o}}$ $80=8 \mathrm{~T}_{\mathrm{o}}$ $\mathrm{T}_{\mathrm{o}}=10^{\circ} \mathrm{C}$
AP EAMCET-24.04.2019
Heat Transfer
149653
One end of a metal rod of length $1.0 \mathrm{~m}$ and area of cross-section $1 \mathrm{~m}^{2}$ is maintained at $100^{\circ} \mathrm{C}$ : If the other end of the rod is maintained at $0^{\circ} \mathrm{C}$, the quantity of heat transmitted through the rod per minute is:
1 $3 \times 10^{3} \mathrm{~J}$
2 $6 \times 10^{3} \mathrm{~J}$
3 $9 \times 10^{3} \mathrm{~J}$
4 $12 \times 10^{3} \mathrm{~J}$
Explanation:
B Given that, Length of $\operatorname{Rod}=1.0 \mathrm{~m}$ Area of cross-section $=1 \mathrm{~m}^{2}$ Temperature $(\Delta t)=100^{\circ} \mathrm{C}$ As we know that, Heat transmitted through the Rod, $\frac{\mathrm{Q}}{\mathrm{t}}=\frac{\mathrm{kA} \Delta \mathrm{t}}{\mathrm{l}}$ $=\frac{1 \times 1 \times(100-0)}{1}$ $\frac{\mathrm{Q}}{60}=100$ $\mathrm{Q}=60 \times 100$ $=6 \times 10^{3}$ Joule
AP EAMCET(Medical)-2000
Heat Transfer
149654
A copper sphere cools from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 10 minutes and to $42^{\circ} \mathrm{C}$ in the next 10 minutes. Calculate the temperature of the surrounding.
1 $18.01^{\circ} \mathrm{C}$
2 $26^{\circ} \mathrm{C}$
3 $10.6^{\circ} \mathrm{C}$
4 $20^{\circ} \mathrm{C}$
Explanation:
B From the Newton's law of cooling. $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right]$ Given, $62^{\circ} \mathrm{C}-50^{\circ} \mathrm{C} \text { in } 10 \mathrm{~min}$ $50^{\circ} \mathrm{C}-42^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$ Then Temperature of surrounding $\theta_{0}=$ ? $\frac{62-50}{10}=K\left[\frac{62+50}{2}-\theta_{0}\right]$ $\frac{50-42}{10}=K\left[\frac{50+42}{2}-\theta_{0}\right]$ Equation (i) / (ii) $\frac{12}{8}=\frac{56-\theta_{0}}{46-\theta_{0}}$ $12 \times 46-12 \theta_{0}=56 \times 8-8 \theta_{0}$ $552-12 \theta_{0}=448-8 \theta_{0}$ $4 \theta_{0}=552-448$ $4 \theta_{0}=104$ $\theta_{0}=\frac{104}{4}$ $\theta_{0}=26^{\circ} \mathrm{C}$
BITSAT-2011
Heat Transfer
149655
A body cools in $7 \mathrm{~min}$ from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. What times (in min) does it take to cool from $40^{\circ} \mathrm{C}$ to $28^{\circ} \mathrm{C}$, if surrounding temperature is $10^{\circ} \mathrm{C}$ ? (Assume Newton's law of cooling)
1 3.5
2 14
3 7
4 10
Explanation:
C Given, $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{Cin} 7 \mathrm{~min}$ $40^{\circ} \mathrm{C}$ to $28^{\circ} \mathrm{C}$ in $\mathrm{t}$ min Given $\mathrm{T}_{\text {surrounding }}=10^{\circ} \mathrm{C}$ Newton's law of cooling $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right)$ Putting the value in equation (i) $\frac{60-40}{7}=\mathrm{K}\left(\frac{60+40}{2}-10\right)$ $\frac{40-28}{\mathrm{t}}=\mathrm{K}\left(\frac{40+28}{2}-10\right)$ Dividing eq $^{\mathrm{n}}$ (ii) / (iii) $\frac{t \times 20}{7 \times 12}=\frac{40}{24}$ $\frac{5 t}{7 \times 3}=\frac{5}{3}$ $t=7 \text { min. }$
149651
A hot container takes $1 \mathrm{~min}$ to cool from $95^{\circ} \mathrm{C}$ to $75^{\circ} \mathrm{C}$. The time it takes to cool from $74^{\circ} \mathrm{C}$ to $54^{\circ} \mathrm{C}$ is [Consider the room temperature as $30^{\circ} \mathrm{C}$ ]
1 $97 \mathrm{~s}$
2 $70 \mathrm{~s}$
3 $102 \mathrm{~s}$
4 $82 \mathrm{~s}$
Explanation:
A Given, initial Temperature $\left(\theta_{1}\right)=95^{\circ} \mathrm{C}$ Final Temperature $\left(\theta_{2}\right)=75^{\circ} \mathrm{C}$ Atmospheric Temperature $(\theta)=30^{\circ} \mathrm{C}$ Time $(\mathrm{t})=1 \mathrm{~min}$ According to Newton's cooling laws $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=-\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta\right)$ $\frac{95-75}{1}=-\mathrm{K}\left(\frac{95+75}{2}-30\right)$ $20=-\mathrm{K}(85-30)$ $\mathrm{K}=-\frac{20}{55}$ Second case, $\theta_{3}=74^{\circ} \mathrm{C}, \theta_{4}=54^{\circ} \mathrm{C}, \theta=30^{\circ} \mathrm{C}$ $\mathrm{K}=-(20 / 55)$ Again use Newton's cooling law, $\frac{74-54}{\mathrm{t}}=-\left(-\frac{20}{55}\right)\left(\frac{74+54}{2}-30\right)$ $\frac{20}{\mathrm{t}}=\frac{20}{55}(64-30)$ $\frac{20}{\mathrm{t}}=\frac{20}{55} \times 34 \Rightarrow \mathrm{t}=\frac{55}{34} \mathrm{~min}$ $=\frac{55}{34} \times 60=97.06 \mathrm{sec} .$
TS EAMCET 30.07.2022
Heat Transfer
149652
A body cools from $60{ }^{\circ} \mathrm{C}$ to $40{ }^{\circ} \mathrm{C}$ in the first 7 minutes and to $28^{\circ} \mathrm{C}$ in the next 7 minutes. The temperature of the surroundings is
1 $10^{\circ} \mathrm{C}$
2 $20^{\circ} \mathrm{C}$
3 $5{ }^{\circ} \mathrm{C}$
4 $30^{\circ} \mathrm{C}$
Explanation:
A From Newton's Law of cooling, $\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{t}}=-\mathrm{K}\left[\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}\right)-\mathrm{T}_{\mathrm{o}}\right]$ $\frac{60-40}{7 \times 60}=-\mathrm{K}\left[\left(\frac{60+40}{2}-\mathrm{T}_{\mathrm{o}}\right)\right.$ $\frac{20}{420}=-\mathrm{K}\left(50-\mathrm{T}_{\mathrm{o}}\right)$ Again cool 7 min, $\frac{40-28}{7 \times 60}=-\mathrm{K}\left[\left(\frac{40+28}{2}\right)-\mathrm{T}_{\mathrm{o}}\right]$ $\frac{12}{420}=-\mathrm{K}\left(34-\mathrm{T}_{\mathrm{o}}\right)$ Dividing Equation (i) by (ii), we get - $\frac{20}{12}=\frac{50-\mathrm{T}_{\mathrm{o}}}{34-\mathrm{T}_{\mathrm{o}}}$ $680-20 \mathrm{~T}_{\mathrm{o}}=600-12 \mathrm{~T}_{\mathrm{o}}$ $80=8 \mathrm{~T}_{\mathrm{o}}$ $\mathrm{T}_{\mathrm{o}}=10^{\circ} \mathrm{C}$
AP EAMCET-24.04.2019
Heat Transfer
149653
One end of a metal rod of length $1.0 \mathrm{~m}$ and area of cross-section $1 \mathrm{~m}^{2}$ is maintained at $100^{\circ} \mathrm{C}$ : If the other end of the rod is maintained at $0^{\circ} \mathrm{C}$, the quantity of heat transmitted through the rod per minute is:
1 $3 \times 10^{3} \mathrm{~J}$
2 $6 \times 10^{3} \mathrm{~J}$
3 $9 \times 10^{3} \mathrm{~J}$
4 $12 \times 10^{3} \mathrm{~J}$
Explanation:
B Given that, Length of $\operatorname{Rod}=1.0 \mathrm{~m}$ Area of cross-section $=1 \mathrm{~m}^{2}$ Temperature $(\Delta t)=100^{\circ} \mathrm{C}$ As we know that, Heat transmitted through the Rod, $\frac{\mathrm{Q}}{\mathrm{t}}=\frac{\mathrm{kA} \Delta \mathrm{t}}{\mathrm{l}}$ $=\frac{1 \times 1 \times(100-0)}{1}$ $\frac{\mathrm{Q}}{60}=100$ $\mathrm{Q}=60 \times 100$ $=6 \times 10^{3}$ Joule
AP EAMCET(Medical)-2000
Heat Transfer
149654
A copper sphere cools from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 10 minutes and to $42^{\circ} \mathrm{C}$ in the next 10 minutes. Calculate the temperature of the surrounding.
1 $18.01^{\circ} \mathrm{C}$
2 $26^{\circ} \mathrm{C}$
3 $10.6^{\circ} \mathrm{C}$
4 $20^{\circ} \mathrm{C}$
Explanation:
B From the Newton's law of cooling. $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right]$ Given, $62^{\circ} \mathrm{C}-50^{\circ} \mathrm{C} \text { in } 10 \mathrm{~min}$ $50^{\circ} \mathrm{C}-42^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$ Then Temperature of surrounding $\theta_{0}=$ ? $\frac{62-50}{10}=K\left[\frac{62+50}{2}-\theta_{0}\right]$ $\frac{50-42}{10}=K\left[\frac{50+42}{2}-\theta_{0}\right]$ Equation (i) / (ii) $\frac{12}{8}=\frac{56-\theta_{0}}{46-\theta_{0}}$ $12 \times 46-12 \theta_{0}=56 \times 8-8 \theta_{0}$ $552-12 \theta_{0}=448-8 \theta_{0}$ $4 \theta_{0}=552-448$ $4 \theta_{0}=104$ $\theta_{0}=\frac{104}{4}$ $\theta_{0}=26^{\circ} \mathrm{C}$
BITSAT-2011
Heat Transfer
149655
A body cools in $7 \mathrm{~min}$ from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. What times (in min) does it take to cool from $40^{\circ} \mathrm{C}$ to $28^{\circ} \mathrm{C}$, if surrounding temperature is $10^{\circ} \mathrm{C}$ ? (Assume Newton's law of cooling)
1 3.5
2 14
3 7
4 10
Explanation:
C Given, $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{Cin} 7 \mathrm{~min}$ $40^{\circ} \mathrm{C}$ to $28^{\circ} \mathrm{C}$ in $\mathrm{t}$ min Given $\mathrm{T}_{\text {surrounding }}=10^{\circ} \mathrm{C}$ Newton's law of cooling $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right)$ Putting the value in equation (i) $\frac{60-40}{7}=\mathrm{K}\left(\frac{60+40}{2}-10\right)$ $\frac{40-28}{\mathrm{t}}=\mathrm{K}\left(\frac{40+28}{2}-10\right)$ Dividing eq $^{\mathrm{n}}$ (ii) / (iii) $\frac{t \times 20}{7 \times 12}=\frac{40}{24}$ $\frac{5 t}{7 \times 3}=\frac{5}{3}$ $t=7 \text { min. }$
