149647
A pan filled with hot food cools from $94^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in $2 \mathrm{~min}$, when the room temperature is $20^{\circ} \mathrm{C}$. The time taken for the food to cool from $86^{\circ} \mathrm{C}$ to $74^{\circ} \mathrm{C}$ will be
1 $500 \mathrm{~s}$
2 $420 \mathrm{~s}$
3 $200 \mathrm{~s}$
4 $210 \mathrm{~s}$
Explanation:
D Given, $\mathrm{T}_{1}=94^{\circ} \mathrm{C}, \mathrm{T}_{2}=86^{\circ} \mathrm{C}, \mathrm{t}=2 \mathrm{~min}$ According to Newton's law of cooling- $\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{2}=\mathrm{K}\left[\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{s}}\right]$ Where, $T_{s}$ is the surrounding temperature which is $20^{\circ} \mathrm{C}$. For first case, $\frac{94-86}{2 \times 60}=\mathrm{K}\left[\frac{94+86}{2}-20\right]$ $\frac{8}{2 \times 60}=\mathrm{K} \times 70$ $\frac{1}{15}=\mathrm{K} \times 70$ For second case, $\frac{86-74}{\mathrm{t}}=\mathrm{K}\left[\frac{86+74}{2}-20\right]$ $\frac{12}{\mathrm{t}}=\mathrm{K} \times 60$ Dividing equation (i) by equation (ii), we get $\frac{\frac{1}{15}}{\frac{12}{t}}=\frac{K \times 70}{K \times 60}$ $\frac{t}{12 \times 15}=\frac{7}{6}$ $t=30 \times 7$ $t=210 s$
AIIMS-27.05.2018(M)]
Heat Transfer
149648
A pan filled with hot food cools from $94^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in $2 \mathrm{~min}$. When the room temperature is $20^{\circ} \mathrm{C}$. How long will it cool from $74^{0} \mathrm{C}$ to $66^{\circ} \mathrm{C}$ ?
1 $2 \mathrm{~min}$
2 $2.8 \mathrm{~min}$
3 $2.5 \mathrm{~min}$
4 $1.8 \mathrm{~min}$
Explanation:
B According to Newton's law of cooling $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{\mathrm{o}}\right]$ For case - (I) $\mathrm{t}=2 \mathrm{~min}$ $\theta_{1}-\theta_{2}=94-86=8^{\circ} \mathrm{C}$ $\theta_{\mathrm{o}}=20^{\circ} \mathrm{C}$ $\frac{\theta_{1}+\theta_{2}}{2}=\frac{94+86}{2}=\frac{180}{2}=90^{\circ} \mathrm{C}$ So, put these value in Newton's law of cooling formula we get $\frac{8}{2}=\mathrm{K}(90-20)$ $\mathrm{K}=\frac{4}{70}$ For case - (II) $\theta_{1}-\theta_{2}=74-66=8^{\circ} \mathrm{C}$ $\mathrm{t}=$ ? $\frac{\theta_{1}+\theta_{2}}{2}=\frac{74+66}{2}=\frac{140}{2}=70^{\circ} \mathrm{C}$ $\theta_{\mathrm{o}}=20^{\circ} \mathrm{C}$ Thus, using Newton's cooling law, $\frac{8}{\mathrm{t}}=\mathrm{K}\left(70^{\circ}-20^{\circ}\right)$ Put the value of $\mathrm{K}$ we get, $\frac{8}{\mathrm{t}}=\frac{4}{70}\left(50^{\circ}\right)$ $\mathrm{t}=\frac{70}{25}=2.8$ Thus, it will take $2.8 \mathrm{~min}$.
Karnataka CET-2016
Heat Transfer
149649
Hot food cools from $94^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in 2 minutes when the room temperature is $20^{\circ} \mathrm{C}$. How long would the food take to cool from $71^{\circ} \mathrm{C}$ to $69^{\circ} \mathrm{C}$ ?
1 $12 \mathrm{sec}$
2 $25 \mathrm{sec}$
3 $16 \mathrm{sec}$
4 $42 \mathrm{sec}$
Explanation:
D We know Newton's law of cooling - $\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{t}}=-\mathrm{K}\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{S}}\right)$ For case-I $\mathrm{T}_{1}=94^{\circ} \mathrm{C}, \mathrm{T}_{2}=86^{\circ} \mathrm{C}, \mathrm{t}=2 \mathrm{~min}$ and $\mathrm{T}_{\mathrm{s}}=20^{\circ} \mathrm{C}$ $\frac{94-86}{2}=-\mathrm{K}\left(\frac{94+86}{2}-20\right)$ $\mathrm{K}=-(4 / 70)=-(2 / 35)$ For case - II $\mathrm{T}_{1}=71^{\circ} \mathrm{C}, \mathrm{T}_{2}=69^{\circ} \mathrm{C}, \mathrm{T}_{\mathrm{s}}=20^{\circ} \mathrm{C}$ and $\mathrm{t}=$ ? From equation (i) $\frac{71-69}{t}=-\left(-\frac{2}{35}\right) \times\left(\frac{71+69}{2}-20\right)$ $t=0.7 \text { min }$ $t=0.7 \times 60$ $t=42 \text { second }$
AMU-2012
Heat Transfer
149650
A body cools from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$. If room temperature is $25^{\circ} \mathrm{C}$ of body at the end of next 10 min, will be
1 $38.5^{\circ} \mathrm{C}$
2 $40^{\circ} \mathrm{C}$
3 $45^{\circ} \mathrm{C}$
4 $42.85^{\circ} \mathrm{C}$
Explanation:
D Given, Initial Temp $\left(\theta_{1}\right)=60^{\circ} \mathrm{C}$, Final Temp. $\left(\theta_{2}\right)=50^{\circ} \mathrm{C}$ Atmospheric Temp. $(\theta)=25^{\circ} \mathrm{C}$ Time $(t)=10$ min According to Newton's law, $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=-\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta\right)$ $\frac{60-50}{10}=-\mathrm{K}\left(\frac{60+50}{2}-25\right)$ $1=-\mathrm{K}(55-25)$ $\mathrm{K}=-\frac{1}{30}$ Case-II Again using Newton's law cooling, $\frac{50-\theta_{3}}{10}=-\left(-\frac{1}{30}\right)\left(\frac{50+\theta_{3}}{2}-25\right)$ $150-3 \theta_{3}=\frac{50+\theta_{3}-50}{2}$ $300-6 \theta_{3}=\theta_{3}$ $7 \theta_{3}=300$ $\theta_{3}=42.85^{\circ} \mathrm{C}$
149647
A pan filled with hot food cools from $94^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in $2 \mathrm{~min}$, when the room temperature is $20^{\circ} \mathrm{C}$. The time taken for the food to cool from $86^{\circ} \mathrm{C}$ to $74^{\circ} \mathrm{C}$ will be
1 $500 \mathrm{~s}$
2 $420 \mathrm{~s}$
3 $200 \mathrm{~s}$
4 $210 \mathrm{~s}$
Explanation:
D Given, $\mathrm{T}_{1}=94^{\circ} \mathrm{C}, \mathrm{T}_{2}=86^{\circ} \mathrm{C}, \mathrm{t}=2 \mathrm{~min}$ According to Newton's law of cooling- $\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{2}=\mathrm{K}\left[\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{s}}\right]$ Where, $T_{s}$ is the surrounding temperature which is $20^{\circ} \mathrm{C}$. For first case, $\frac{94-86}{2 \times 60}=\mathrm{K}\left[\frac{94+86}{2}-20\right]$ $\frac{8}{2 \times 60}=\mathrm{K} \times 70$ $\frac{1}{15}=\mathrm{K} \times 70$ For second case, $\frac{86-74}{\mathrm{t}}=\mathrm{K}\left[\frac{86+74}{2}-20\right]$ $\frac{12}{\mathrm{t}}=\mathrm{K} \times 60$ Dividing equation (i) by equation (ii), we get $\frac{\frac{1}{15}}{\frac{12}{t}}=\frac{K \times 70}{K \times 60}$ $\frac{t}{12 \times 15}=\frac{7}{6}$ $t=30 \times 7$ $t=210 s$
AIIMS-27.05.2018(M)]
Heat Transfer
149648
A pan filled with hot food cools from $94^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in $2 \mathrm{~min}$. When the room temperature is $20^{\circ} \mathrm{C}$. How long will it cool from $74^{0} \mathrm{C}$ to $66^{\circ} \mathrm{C}$ ?
1 $2 \mathrm{~min}$
2 $2.8 \mathrm{~min}$
3 $2.5 \mathrm{~min}$
4 $1.8 \mathrm{~min}$
Explanation:
B According to Newton's law of cooling $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{\mathrm{o}}\right]$ For case - (I) $\mathrm{t}=2 \mathrm{~min}$ $\theta_{1}-\theta_{2}=94-86=8^{\circ} \mathrm{C}$ $\theta_{\mathrm{o}}=20^{\circ} \mathrm{C}$ $\frac{\theta_{1}+\theta_{2}}{2}=\frac{94+86}{2}=\frac{180}{2}=90^{\circ} \mathrm{C}$ So, put these value in Newton's law of cooling formula we get $\frac{8}{2}=\mathrm{K}(90-20)$ $\mathrm{K}=\frac{4}{70}$ For case - (II) $\theta_{1}-\theta_{2}=74-66=8^{\circ} \mathrm{C}$ $\mathrm{t}=$ ? $\frac{\theta_{1}+\theta_{2}}{2}=\frac{74+66}{2}=\frac{140}{2}=70^{\circ} \mathrm{C}$ $\theta_{\mathrm{o}}=20^{\circ} \mathrm{C}$ Thus, using Newton's cooling law, $\frac{8}{\mathrm{t}}=\mathrm{K}\left(70^{\circ}-20^{\circ}\right)$ Put the value of $\mathrm{K}$ we get, $\frac{8}{\mathrm{t}}=\frac{4}{70}\left(50^{\circ}\right)$ $\mathrm{t}=\frac{70}{25}=2.8$ Thus, it will take $2.8 \mathrm{~min}$.
Karnataka CET-2016
Heat Transfer
149649
Hot food cools from $94^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in 2 minutes when the room temperature is $20^{\circ} \mathrm{C}$. How long would the food take to cool from $71^{\circ} \mathrm{C}$ to $69^{\circ} \mathrm{C}$ ?
1 $12 \mathrm{sec}$
2 $25 \mathrm{sec}$
3 $16 \mathrm{sec}$
4 $42 \mathrm{sec}$
Explanation:
D We know Newton's law of cooling - $\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{t}}=-\mathrm{K}\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{S}}\right)$ For case-I $\mathrm{T}_{1}=94^{\circ} \mathrm{C}, \mathrm{T}_{2}=86^{\circ} \mathrm{C}, \mathrm{t}=2 \mathrm{~min}$ and $\mathrm{T}_{\mathrm{s}}=20^{\circ} \mathrm{C}$ $\frac{94-86}{2}=-\mathrm{K}\left(\frac{94+86}{2}-20\right)$ $\mathrm{K}=-(4 / 70)=-(2 / 35)$ For case - II $\mathrm{T}_{1}=71^{\circ} \mathrm{C}, \mathrm{T}_{2}=69^{\circ} \mathrm{C}, \mathrm{T}_{\mathrm{s}}=20^{\circ} \mathrm{C}$ and $\mathrm{t}=$ ? From equation (i) $\frac{71-69}{t}=-\left(-\frac{2}{35}\right) \times\left(\frac{71+69}{2}-20\right)$ $t=0.7 \text { min }$ $t=0.7 \times 60$ $t=42 \text { second }$
AMU-2012
Heat Transfer
149650
A body cools from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$. If room temperature is $25^{\circ} \mathrm{C}$ of body at the end of next 10 min, will be
1 $38.5^{\circ} \mathrm{C}$
2 $40^{\circ} \mathrm{C}$
3 $45^{\circ} \mathrm{C}$
4 $42.85^{\circ} \mathrm{C}$
Explanation:
D Given, Initial Temp $\left(\theta_{1}\right)=60^{\circ} \mathrm{C}$, Final Temp. $\left(\theta_{2}\right)=50^{\circ} \mathrm{C}$ Atmospheric Temp. $(\theta)=25^{\circ} \mathrm{C}$ Time $(t)=10$ min According to Newton's law, $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=-\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta\right)$ $\frac{60-50}{10}=-\mathrm{K}\left(\frac{60+50}{2}-25\right)$ $1=-\mathrm{K}(55-25)$ $\mathrm{K}=-\frac{1}{30}$ Case-II Again using Newton's law cooling, $\frac{50-\theta_{3}}{10}=-\left(-\frac{1}{30}\right)\left(\frac{50+\theta_{3}}{2}-25\right)$ $150-3 \theta_{3}=\frac{50+\theta_{3}-50}{2}$ $300-6 \theta_{3}=\theta_{3}$ $7 \theta_{3}=300$ $\theta_{3}=42.85^{\circ} \mathrm{C}$
149647
A pan filled with hot food cools from $94^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in $2 \mathrm{~min}$, when the room temperature is $20^{\circ} \mathrm{C}$. The time taken for the food to cool from $86^{\circ} \mathrm{C}$ to $74^{\circ} \mathrm{C}$ will be
1 $500 \mathrm{~s}$
2 $420 \mathrm{~s}$
3 $200 \mathrm{~s}$
4 $210 \mathrm{~s}$
Explanation:
D Given, $\mathrm{T}_{1}=94^{\circ} \mathrm{C}, \mathrm{T}_{2}=86^{\circ} \mathrm{C}, \mathrm{t}=2 \mathrm{~min}$ According to Newton's law of cooling- $\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{2}=\mathrm{K}\left[\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{s}}\right]$ Where, $T_{s}$ is the surrounding temperature which is $20^{\circ} \mathrm{C}$. For first case, $\frac{94-86}{2 \times 60}=\mathrm{K}\left[\frac{94+86}{2}-20\right]$ $\frac{8}{2 \times 60}=\mathrm{K} \times 70$ $\frac{1}{15}=\mathrm{K} \times 70$ For second case, $\frac{86-74}{\mathrm{t}}=\mathrm{K}\left[\frac{86+74}{2}-20\right]$ $\frac{12}{\mathrm{t}}=\mathrm{K} \times 60$ Dividing equation (i) by equation (ii), we get $\frac{\frac{1}{15}}{\frac{12}{t}}=\frac{K \times 70}{K \times 60}$ $\frac{t}{12 \times 15}=\frac{7}{6}$ $t=30 \times 7$ $t=210 s$
AIIMS-27.05.2018(M)]
Heat Transfer
149648
A pan filled with hot food cools from $94^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in $2 \mathrm{~min}$. When the room temperature is $20^{\circ} \mathrm{C}$. How long will it cool from $74^{0} \mathrm{C}$ to $66^{\circ} \mathrm{C}$ ?
1 $2 \mathrm{~min}$
2 $2.8 \mathrm{~min}$
3 $2.5 \mathrm{~min}$
4 $1.8 \mathrm{~min}$
Explanation:
B According to Newton's law of cooling $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{\mathrm{o}}\right]$ For case - (I) $\mathrm{t}=2 \mathrm{~min}$ $\theta_{1}-\theta_{2}=94-86=8^{\circ} \mathrm{C}$ $\theta_{\mathrm{o}}=20^{\circ} \mathrm{C}$ $\frac{\theta_{1}+\theta_{2}}{2}=\frac{94+86}{2}=\frac{180}{2}=90^{\circ} \mathrm{C}$ So, put these value in Newton's law of cooling formula we get $\frac{8}{2}=\mathrm{K}(90-20)$ $\mathrm{K}=\frac{4}{70}$ For case - (II) $\theta_{1}-\theta_{2}=74-66=8^{\circ} \mathrm{C}$ $\mathrm{t}=$ ? $\frac{\theta_{1}+\theta_{2}}{2}=\frac{74+66}{2}=\frac{140}{2}=70^{\circ} \mathrm{C}$ $\theta_{\mathrm{o}}=20^{\circ} \mathrm{C}$ Thus, using Newton's cooling law, $\frac{8}{\mathrm{t}}=\mathrm{K}\left(70^{\circ}-20^{\circ}\right)$ Put the value of $\mathrm{K}$ we get, $\frac{8}{\mathrm{t}}=\frac{4}{70}\left(50^{\circ}\right)$ $\mathrm{t}=\frac{70}{25}=2.8$ Thus, it will take $2.8 \mathrm{~min}$.
Karnataka CET-2016
Heat Transfer
149649
Hot food cools from $94^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in 2 minutes when the room temperature is $20^{\circ} \mathrm{C}$. How long would the food take to cool from $71^{\circ} \mathrm{C}$ to $69^{\circ} \mathrm{C}$ ?
1 $12 \mathrm{sec}$
2 $25 \mathrm{sec}$
3 $16 \mathrm{sec}$
4 $42 \mathrm{sec}$
Explanation:
D We know Newton's law of cooling - $\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{t}}=-\mathrm{K}\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{S}}\right)$ For case-I $\mathrm{T}_{1}=94^{\circ} \mathrm{C}, \mathrm{T}_{2}=86^{\circ} \mathrm{C}, \mathrm{t}=2 \mathrm{~min}$ and $\mathrm{T}_{\mathrm{s}}=20^{\circ} \mathrm{C}$ $\frac{94-86}{2}=-\mathrm{K}\left(\frac{94+86}{2}-20\right)$ $\mathrm{K}=-(4 / 70)=-(2 / 35)$ For case - II $\mathrm{T}_{1}=71^{\circ} \mathrm{C}, \mathrm{T}_{2}=69^{\circ} \mathrm{C}, \mathrm{T}_{\mathrm{s}}=20^{\circ} \mathrm{C}$ and $\mathrm{t}=$ ? From equation (i) $\frac{71-69}{t}=-\left(-\frac{2}{35}\right) \times\left(\frac{71+69}{2}-20\right)$ $t=0.7 \text { min }$ $t=0.7 \times 60$ $t=42 \text { second }$
AMU-2012
Heat Transfer
149650
A body cools from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$. If room temperature is $25^{\circ} \mathrm{C}$ of body at the end of next 10 min, will be
1 $38.5^{\circ} \mathrm{C}$
2 $40^{\circ} \mathrm{C}$
3 $45^{\circ} \mathrm{C}$
4 $42.85^{\circ} \mathrm{C}$
Explanation:
D Given, Initial Temp $\left(\theta_{1}\right)=60^{\circ} \mathrm{C}$, Final Temp. $\left(\theta_{2}\right)=50^{\circ} \mathrm{C}$ Atmospheric Temp. $(\theta)=25^{\circ} \mathrm{C}$ Time $(t)=10$ min According to Newton's law, $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=-\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta\right)$ $\frac{60-50}{10}=-\mathrm{K}\left(\frac{60+50}{2}-25\right)$ $1=-\mathrm{K}(55-25)$ $\mathrm{K}=-\frac{1}{30}$ Case-II Again using Newton's law cooling, $\frac{50-\theta_{3}}{10}=-\left(-\frac{1}{30}\right)\left(\frac{50+\theta_{3}}{2}-25\right)$ $150-3 \theta_{3}=\frac{50+\theta_{3}-50}{2}$ $300-6 \theta_{3}=\theta_{3}$ $7 \theta_{3}=300$ $\theta_{3}=42.85^{\circ} \mathrm{C}$
149647
A pan filled with hot food cools from $94^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in $2 \mathrm{~min}$, when the room temperature is $20^{\circ} \mathrm{C}$. The time taken for the food to cool from $86^{\circ} \mathrm{C}$ to $74^{\circ} \mathrm{C}$ will be
1 $500 \mathrm{~s}$
2 $420 \mathrm{~s}$
3 $200 \mathrm{~s}$
4 $210 \mathrm{~s}$
Explanation:
D Given, $\mathrm{T}_{1}=94^{\circ} \mathrm{C}, \mathrm{T}_{2}=86^{\circ} \mathrm{C}, \mathrm{t}=2 \mathrm{~min}$ According to Newton's law of cooling- $\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{2}=\mathrm{K}\left[\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{s}}\right]$ Where, $T_{s}$ is the surrounding temperature which is $20^{\circ} \mathrm{C}$. For first case, $\frac{94-86}{2 \times 60}=\mathrm{K}\left[\frac{94+86}{2}-20\right]$ $\frac{8}{2 \times 60}=\mathrm{K} \times 70$ $\frac{1}{15}=\mathrm{K} \times 70$ For second case, $\frac{86-74}{\mathrm{t}}=\mathrm{K}\left[\frac{86+74}{2}-20\right]$ $\frac{12}{\mathrm{t}}=\mathrm{K} \times 60$ Dividing equation (i) by equation (ii), we get $\frac{\frac{1}{15}}{\frac{12}{t}}=\frac{K \times 70}{K \times 60}$ $\frac{t}{12 \times 15}=\frac{7}{6}$ $t=30 \times 7$ $t=210 s$
AIIMS-27.05.2018(M)]
Heat Transfer
149648
A pan filled with hot food cools from $94^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in $2 \mathrm{~min}$. When the room temperature is $20^{\circ} \mathrm{C}$. How long will it cool from $74^{0} \mathrm{C}$ to $66^{\circ} \mathrm{C}$ ?
1 $2 \mathrm{~min}$
2 $2.8 \mathrm{~min}$
3 $2.5 \mathrm{~min}$
4 $1.8 \mathrm{~min}$
Explanation:
B According to Newton's law of cooling $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{\mathrm{o}}\right]$ For case - (I) $\mathrm{t}=2 \mathrm{~min}$ $\theta_{1}-\theta_{2}=94-86=8^{\circ} \mathrm{C}$ $\theta_{\mathrm{o}}=20^{\circ} \mathrm{C}$ $\frac{\theta_{1}+\theta_{2}}{2}=\frac{94+86}{2}=\frac{180}{2}=90^{\circ} \mathrm{C}$ So, put these value in Newton's law of cooling formula we get $\frac{8}{2}=\mathrm{K}(90-20)$ $\mathrm{K}=\frac{4}{70}$ For case - (II) $\theta_{1}-\theta_{2}=74-66=8^{\circ} \mathrm{C}$ $\mathrm{t}=$ ? $\frac{\theta_{1}+\theta_{2}}{2}=\frac{74+66}{2}=\frac{140}{2}=70^{\circ} \mathrm{C}$ $\theta_{\mathrm{o}}=20^{\circ} \mathrm{C}$ Thus, using Newton's cooling law, $\frac{8}{\mathrm{t}}=\mathrm{K}\left(70^{\circ}-20^{\circ}\right)$ Put the value of $\mathrm{K}$ we get, $\frac{8}{\mathrm{t}}=\frac{4}{70}\left(50^{\circ}\right)$ $\mathrm{t}=\frac{70}{25}=2.8$ Thus, it will take $2.8 \mathrm{~min}$.
Karnataka CET-2016
Heat Transfer
149649
Hot food cools from $94^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in 2 minutes when the room temperature is $20^{\circ} \mathrm{C}$. How long would the food take to cool from $71^{\circ} \mathrm{C}$ to $69^{\circ} \mathrm{C}$ ?
1 $12 \mathrm{sec}$
2 $25 \mathrm{sec}$
3 $16 \mathrm{sec}$
4 $42 \mathrm{sec}$
Explanation:
D We know Newton's law of cooling - $\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{t}}=-\mathrm{K}\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{S}}\right)$ For case-I $\mathrm{T}_{1}=94^{\circ} \mathrm{C}, \mathrm{T}_{2}=86^{\circ} \mathrm{C}, \mathrm{t}=2 \mathrm{~min}$ and $\mathrm{T}_{\mathrm{s}}=20^{\circ} \mathrm{C}$ $\frac{94-86}{2}=-\mathrm{K}\left(\frac{94+86}{2}-20\right)$ $\mathrm{K}=-(4 / 70)=-(2 / 35)$ For case - II $\mathrm{T}_{1}=71^{\circ} \mathrm{C}, \mathrm{T}_{2}=69^{\circ} \mathrm{C}, \mathrm{T}_{\mathrm{s}}=20^{\circ} \mathrm{C}$ and $\mathrm{t}=$ ? From equation (i) $\frac{71-69}{t}=-\left(-\frac{2}{35}\right) \times\left(\frac{71+69}{2}-20\right)$ $t=0.7 \text { min }$ $t=0.7 \times 60$ $t=42 \text { second }$
AMU-2012
Heat Transfer
149650
A body cools from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$. If room temperature is $25^{\circ} \mathrm{C}$ of body at the end of next 10 min, will be
1 $38.5^{\circ} \mathrm{C}$
2 $40^{\circ} \mathrm{C}$
3 $45^{\circ} \mathrm{C}$
4 $42.85^{\circ} \mathrm{C}$
Explanation:
D Given, Initial Temp $\left(\theta_{1}\right)=60^{\circ} \mathrm{C}$, Final Temp. $\left(\theta_{2}\right)=50^{\circ} \mathrm{C}$ Atmospheric Temp. $(\theta)=25^{\circ} \mathrm{C}$ Time $(t)=10$ min According to Newton's law, $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=-\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta\right)$ $\frac{60-50}{10}=-\mathrm{K}\left(\frac{60+50}{2}-25\right)$ $1=-\mathrm{K}(55-25)$ $\mathrm{K}=-\frac{1}{30}$ Case-II Again using Newton's law cooling, $\frac{50-\theta_{3}}{10}=-\left(-\frac{1}{30}\right)\left(\frac{50+\theta_{3}}{2}-25\right)$ $150-3 \theta_{3}=\frac{50+\theta_{3}-50}{2}$ $300-6 \theta_{3}=\theta_{3}$ $7 \theta_{3}=300$ $\theta_{3}=42.85^{\circ} \mathrm{C}$
