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PHXI11:THERMAL PROPERTIES OF MATTER

366657 A body cools from $80^{\circ} C$ to $64^{\circ} C$ in $5 \min$ and same body cools from $80^{\circ} C$ to $52^{\circ} C$ in $10 \min$ , what is the temperature of the surrounding?

1

24^{\circ} C

2

28^{\circ} C

3

22^{\circ} C

4

25^{\circ} C

Explanation:

Let $T$ be the temperature of the surrounding. According to Newton's law of cooling,
$\frac{80 - 64}{5} = k [\frac{80 + 64}{2} - T] (1)$
$a n d \frac{80 - 52}{10} = k [\frac{80 + 52}{2} - T] (2)$
Solving Eqs.(1) and (2), we get
$T = 24^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366658 In a room where temperature is $30^{\circ} C$ , a body cools from $61^{\circ} C$ to $59^{\circ} C$ in 4 minutes. The time taken by the body to cool from $51^{\circ} C$ to $49^{\circ} C$ under the same conditions would be

1 4 minutes

2 6 minutes

3 5 minutes

4 8 minutes

Explanation:

From Newton's cooling law, $\frac{(T_{1} - T_{2})}{t} = k [\frac{T_{1} + T_{2}}{2} - T_{0}]$
For first case, $T_{1} = 61^{\circ} C, T_{2} = 59^{\circ} C, T_{0} = 30^{\circ} C, t = 4$ minutes
$∴ \frac{61 - 59}{4} = k [\frac{61 + 59}{2} - 30] \Rightarrow k = \frac{1}{60}$
For second case, $T_{1} = 51^{\circ} C, T_{2} = 49^{\circ} C, t = ?, T_{0} = 30^{\circ} C$
$∴ [\frac{51 - 49}{t}] = k [\frac{51 + 49}{2} - 30]$
$\frac{2}{t} = k 20 [∵ k = \frac{1}{60}]$
$\Rightarrow t = \frac{2 \times 60}{20} = 6$ minutes.
So correct option is (2)

PHXI11:THERMAL PROPERTIES OF MATTER

366659 Hot water cools from $60^{\circ} C$ to $50^{\circ} C$ in the first 10 minutes and to $42^{\circ} C$ in the next 10 minutes. The temperature of the surroundings is:

1

25^{\circ} C

2

10^{\circ} C

3

15^{\circ} C

4

20^{\circ} C

Explanation:

By Newton's law of cooling
$\frac{θ_{1} - θ_{2}}{t} = - K [\frac{θ_{1} + θ_{2}}{2} - θ_{0}]$
where $θ_{0}$ is the temperature of surrounding.
Now, hot water cools from $60^{\circ} C$ to $50^{\circ} C$ in 10 minutes,
$\frac{60 - 50}{10} = - K [\frac{60 + 50}{2} - θ_{0}] (1)$
Again, it cools from $50^{\circ} C$ to $42^{\circ} C$ in next 10 minutes.
$\frac{50 - 42}{10} = - K [\frac{50 + 42}{2} - θ_{0}] (2)$
Dividing equation (1) by (2) we get
$\Rightarrow θ_{0} = 10^{\circ} C$

JEE - 2014

PHXI11:THERMAL PROPERTIES OF MATTER

366660 A cup of coffee cools from $90^{\circ} C$ to $80^{\circ} C$ in $t$ minutes, when the room temperature is $20^{\circ} C$ . The time taken by a similar cup of coffee to cool from $80^{\circ} C$ to $60^{\circ} C$ at a room temperature same at $20^{\circ} C$ is

1

\frac{13}{5} t

2

\frac{10}{13} t

3

\frac{5}{13} t

4

\frac{13}{10} t

Explanation:

From newton's laws of cooling
$\begin{array}{r} \frac{T_{1} - T_{2}}{Δ t} = k [\frac{T_{1} + T_{2}}{2} - T_{s}] \\ \frac{90 - 80}{t} = k [\frac{90 + 80}{2} - 20] \\ \frac{80 - 60}{t^{'}} = k [\frac{80 + 60}{2} - 20] \\ \Rightarrow \frac{10}{t} \times \frac{t^{'}}{20} = \frac{65}{50} \Rightarrow t^{'} = \frac{13}{5} t \end{array}$

NEET - 2021

PHXI11:THERMAL PROPERTIES OF MATTER

366657 A body cools from $80^{\circ} C$ to $64^{\circ} C$ in $5 \min$ and same body cools from $80^{\circ} C$ to $52^{\circ} C$ in $10 \min$ , what is the temperature of the surrounding?

1

24^{\circ} C

2

28^{\circ} C

3

22^{\circ} C

4

25^{\circ} C

Explanation:

Let $T$ be the temperature of the surrounding. According to Newton's law of cooling,
$\frac{80 - 64}{5} = k [\frac{80 + 64}{2} - T] (1)$
$a n d \frac{80 - 52}{10} = k [\frac{80 + 52}{2} - T] (2)$
Solving Eqs.(1) and (2), we get
$T = 24^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366658 In a room where temperature is $30^{\circ} C$ , a body cools from $61^{\circ} C$ to $59^{\circ} C$ in 4 minutes. The time taken by the body to cool from $51^{\circ} C$ to $49^{\circ} C$ under the same conditions would be

1 4 minutes

2 6 minutes

3 5 minutes

4 8 minutes

Explanation:

From Newton's cooling law, $\frac{(T_{1} - T_{2})}{t} = k [\frac{T_{1} + T_{2}}{2} - T_{0}]$
For first case, $T_{1} = 61^{\circ} C, T_{2} = 59^{\circ} C, T_{0} = 30^{\circ} C, t = 4$ minutes
$∴ \frac{61 - 59}{4} = k [\frac{61 + 59}{2} - 30] \Rightarrow k = \frac{1}{60}$
For second case, $T_{1} = 51^{\circ} C, T_{2} = 49^{\circ} C, t = ?, T_{0} = 30^{\circ} C$
$∴ [\frac{51 - 49}{t}] = k [\frac{51 + 49}{2} - 30]$
$\frac{2}{t} = k 20 [∵ k = \frac{1}{60}]$
$\Rightarrow t = \frac{2 \times 60}{20} = 6$ minutes.
So correct option is (2)

PHXI11:THERMAL PROPERTIES OF MATTER

366659 Hot water cools from $60^{\circ} C$ to $50^{\circ} C$ in the first 10 minutes and to $42^{\circ} C$ in the next 10 minutes. The temperature of the surroundings is:

1

25^{\circ} C

2

10^{\circ} C

3

15^{\circ} C

4

20^{\circ} C

Explanation:

By Newton's law of cooling
$\frac{θ_{1} - θ_{2}}{t} = - K [\frac{θ_{1} + θ_{2}}{2} - θ_{0}]$
where $θ_{0}$ is the temperature of surrounding.
Now, hot water cools from $60^{\circ} C$ to $50^{\circ} C$ in 10 minutes,
$\frac{60 - 50}{10} = - K [\frac{60 + 50}{2} - θ_{0}] (1)$
Again, it cools from $50^{\circ} C$ to $42^{\circ} C$ in next 10 minutes.
$\frac{50 - 42}{10} = - K [\frac{50 + 42}{2} - θ_{0}] (2)$
Dividing equation (1) by (2) we get
$\Rightarrow θ_{0} = 10^{\circ} C$

JEE - 2014

PHXI11:THERMAL PROPERTIES OF MATTER

366660 A cup of coffee cools from $90^{\circ} C$ to $80^{\circ} C$ in $t$ minutes, when the room temperature is $20^{\circ} C$ . The time taken by a similar cup of coffee to cool from $80^{\circ} C$ to $60^{\circ} C$ at a room temperature same at $20^{\circ} C$ is

1

\frac{13}{5} t

2

\frac{10}{13} t

3

\frac{5}{13} t

4

\frac{13}{10} t

Explanation:

From newton's laws of cooling
$\begin{array}{r} \frac{T_{1} - T_{2}}{Δ t} = k [\frac{T_{1} + T_{2}}{2} - T_{s}] \\ \frac{90 - 80}{t} = k [\frac{90 + 80}{2} - 20] \\ \frac{80 - 60}{t^{'}} = k [\frac{80 + 60}{2} - 20] \\ \Rightarrow \frac{10}{t} \times \frac{t^{'}}{20} = \frac{65}{50} \Rightarrow t^{'} = \frac{13}{5} t \end{array}$

NEET - 2021

PHXI11:THERMAL PROPERTIES OF MATTER

366657 A body cools from $80^{\circ} C$ to $64^{\circ} C$ in $5 \min$ and same body cools from $80^{\circ} C$ to $52^{\circ} C$ in $10 \min$ , what is the temperature of the surrounding?

1

24^{\circ} C

2

28^{\circ} C

3

22^{\circ} C

4

25^{\circ} C

Explanation:

Let $T$ be the temperature of the surrounding. According to Newton's law of cooling,
$\frac{80 - 64}{5} = k [\frac{80 + 64}{2} - T] (1)$
$a n d \frac{80 - 52}{10} = k [\frac{80 + 52}{2} - T] (2)$
Solving Eqs.(1) and (2), we get
$T = 24^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366658 In a room where temperature is $30^{\circ} C$ , a body cools from $61^{\circ} C$ to $59^{\circ} C$ in 4 minutes. The time taken by the body to cool from $51^{\circ} C$ to $49^{\circ} C$ under the same conditions would be

1 4 minutes

2 6 minutes

3 5 minutes

4 8 minutes

Explanation:

From Newton's cooling law, $\frac{(T_{1} - T_{2})}{t} = k [\frac{T_{1} + T_{2}}{2} - T_{0}]$
For first case, $T_{1} = 61^{\circ} C, T_{2} = 59^{\circ} C, T_{0} = 30^{\circ} C, t = 4$ minutes
$∴ \frac{61 - 59}{4} = k [\frac{61 + 59}{2} - 30] \Rightarrow k = \frac{1}{60}$
For second case, $T_{1} = 51^{\circ} C, T_{2} = 49^{\circ} C, t = ?, T_{0} = 30^{\circ} C$
$∴ [\frac{51 - 49}{t}] = k [\frac{51 + 49}{2} - 30]$
$\frac{2}{t} = k 20 [∵ k = \frac{1}{60}]$
$\Rightarrow t = \frac{2 \times 60}{20} = 6$ minutes.
So correct option is (2)

PHXI11:THERMAL PROPERTIES OF MATTER

366659 Hot water cools from $60^{\circ} C$ to $50^{\circ} C$ in the first 10 minutes and to $42^{\circ} C$ in the next 10 minutes. The temperature of the surroundings is:

1

25^{\circ} C

2

10^{\circ} C

3

15^{\circ} C

4

20^{\circ} C

Explanation:

By Newton's law of cooling
$\frac{θ_{1} - θ_{2}}{t} = - K [\frac{θ_{1} + θ_{2}}{2} - θ_{0}]$
where $θ_{0}$ is the temperature of surrounding.
Now, hot water cools from $60^{\circ} C$ to $50^{\circ} C$ in 10 minutes,
$\frac{60 - 50}{10} = - K [\frac{60 + 50}{2} - θ_{0}] (1)$
Again, it cools from $50^{\circ} C$ to $42^{\circ} C$ in next 10 minutes.
$\frac{50 - 42}{10} = - K [\frac{50 + 42}{2} - θ_{0}] (2)$
Dividing equation (1) by (2) we get
$\Rightarrow θ_{0} = 10^{\circ} C$

JEE - 2014

PHXI11:THERMAL PROPERTIES OF MATTER

366660 A cup of coffee cools from $90^{\circ} C$ to $80^{\circ} C$ in $t$ minutes, when the room temperature is $20^{\circ} C$ . The time taken by a similar cup of coffee to cool from $80^{\circ} C$ to $60^{\circ} C$ at a room temperature same at $20^{\circ} C$ is

1

\frac{13}{5} t

2

\frac{10}{13} t

3

\frac{5}{13} t

4

\frac{13}{10} t

Explanation:

From newton's laws of cooling
$\begin{array}{r} \frac{T_{1} - T_{2}}{Δ t} = k [\frac{T_{1} + T_{2}}{2} - T_{s}] \\ \frac{90 - 80}{t} = k [\frac{90 + 80}{2} - 20] \\ \frac{80 - 60}{t^{'}} = k [\frac{80 + 60}{2} - 20] \\ \Rightarrow \frac{10}{t} \times \frac{t^{'}}{20} = \frac{65}{50} \Rightarrow t^{'} = \frac{13}{5} t \end{array}$

NEET - 2021

PHXI11:THERMAL PROPERTIES OF MATTER

366657 A body cools from $80^{\circ} C$ to $64^{\circ} C$ in $5 \min$ and same body cools from $80^{\circ} C$ to $52^{\circ} C$ in $10 \min$ , what is the temperature of the surrounding?

1

24^{\circ} C

2

28^{\circ} C

3

22^{\circ} C

4

25^{\circ} C

Explanation:

Let $T$ be the temperature of the surrounding. According to Newton's law of cooling,
$\frac{80 - 64}{5} = k [\frac{80 + 64}{2} - T] (1)$
$a n d \frac{80 - 52}{10} = k [\frac{80 + 52}{2} - T] (2)$
Solving Eqs.(1) and (2), we get
$T = 24^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366658 In a room where temperature is $30^{\circ} C$ , a body cools from $61^{\circ} C$ to $59^{\circ} C$ in 4 minutes. The time taken by the body to cool from $51^{\circ} C$ to $49^{\circ} C$ under the same conditions would be

1 4 minutes

2 6 minutes

3 5 minutes

4 8 minutes

Explanation:

From Newton's cooling law, $\frac{(T_{1} - T_{2})}{t} = k [\frac{T_{1} + T_{2}}{2} - T_{0}]$
For first case, $T_{1} = 61^{\circ} C, T_{2} = 59^{\circ} C, T_{0} = 30^{\circ} C, t = 4$ minutes
$∴ \frac{61 - 59}{4} = k [\frac{61 + 59}{2} - 30] \Rightarrow k = \frac{1}{60}$
For second case, $T_{1} = 51^{\circ} C, T_{2} = 49^{\circ} C, t = ?, T_{0} = 30^{\circ} C$
$∴ [\frac{51 - 49}{t}] = k [\frac{51 + 49}{2} - 30]$
$\frac{2}{t} = k 20 [∵ k = \frac{1}{60}]$
$\Rightarrow t = \frac{2 \times 60}{20} = 6$ minutes.
So correct option is (2)

PHXI11:THERMAL PROPERTIES OF MATTER

366659 Hot water cools from $60^{\circ} C$ to $50^{\circ} C$ in the first 10 minutes and to $42^{\circ} C$ in the next 10 minutes. The temperature of the surroundings is:

1

25^{\circ} C

2

10^{\circ} C

3

15^{\circ} C

4

20^{\circ} C

Explanation:

By Newton's law of cooling
$\frac{θ_{1} - θ_{2}}{t} = - K [\frac{θ_{1} + θ_{2}}{2} - θ_{0}]$
where $θ_{0}$ is the temperature of surrounding.
Now, hot water cools from $60^{\circ} C$ to $50^{\circ} C$ in 10 minutes,
$\frac{60 - 50}{10} = - K [\frac{60 + 50}{2} - θ_{0}] (1)$
Again, it cools from $50^{\circ} C$ to $42^{\circ} C$ in next 10 minutes.
$\frac{50 - 42}{10} = - K [\frac{50 + 42}{2} - θ_{0}] (2)$
Dividing equation (1) by (2) we get
$\Rightarrow θ_{0} = 10^{\circ} C$

JEE - 2014

PHXI11:THERMAL PROPERTIES OF MATTER

366660 A cup of coffee cools from $90^{\circ} C$ to $80^{\circ} C$ in $t$ minutes, when the room temperature is $20^{\circ} C$ . The time taken by a similar cup of coffee to cool from $80^{\circ} C$ to $60^{\circ} C$ at a room temperature same at $20^{\circ} C$ is

1

\frac{13}{5} t

2

\frac{10}{13} t

3

\frac{5}{13} t

4

\frac{13}{10} t

Explanation:

From newton's laws of cooling
$\begin{array}{r} \frac{T_{1} - T_{2}}{Δ t} = k [\frac{T_{1} + T_{2}}{2} - T_{s}] \\ \frac{90 - 80}{t} = k [\frac{90 + 80}{2} - 20] \\ \frac{80 - 60}{t^{'}} = k [\frac{80 + 60}{2} - 20] \\ \Rightarrow \frac{10}{t} \times \frac{t^{'}}{20} = \frac{65}{50} \Rightarrow t^{'} = \frac{13}{5} t \end{array}$

NEET - 2021

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