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PHXI11:THERMAL PROPERTIES OF MATTER

366639 A body cools from $80^{\circ} C$ to $50^{\circ} C$ in $5 min$ . Calculate the time it takes to cool from $60^{\circ} C$ to $30^{\circ} C$ . The temperature of the surroundings is $20^{\circ} C$ .

1

9 \min

2

7 \min

3

8 \min

4

10 \min

Explanation:

Given, initial temperature $T_{1} = 80^{\circ} C$
Final temperature $T_{2} = 50^{\circ} C$
Temperature of the surroundings $T_{0} = 20^{\circ} C$
$t_{1} = 5 min$
According to Newton's law of cooling.
$\begin{aligned} Rate of cooling, \frac{d T}{d t} = k [\frac{T_{1} + T_{2}}{2} - T_{0}] \\ \frac{(80 - 50)}{5} = k (\frac{80 + 50}{2} - 20) \\ \frac{30}{5} = k (65 - 20) \\ \Rightarrow k = \frac{6}{45} = \frac{2}{15} \end{aligned}$
In second case,
Initial temperature $T_{1}^{'} = 60^{\circ} C$
Final temperature $T_{2}^{'} = 30^{\circ} C$
$t^{'} = ?$
$N o w, \frac{(60 - 30)}{t^{'}} = \frac{2}{15} (\frac{60 + 30}{2} - 20)$
$\Rightarrow t^{'} = \frac{30 \times 15}{2 \times 25} = 9 min .$

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366640 A body cools from $80^{\circ} C$ to $60^{\circ} C$ in $10 \min$ , when the temperature of surroundings is $30^{\circ} C$ . The temperature of the body after next $10 min$ will be

1

40^{\circ} C

2

43^{\circ} C

3

48^{\circ} C

4

52^{\circ} C

Explanation:

From Newton's law of cooling
$\frac{(80 - 60)}{10} = K (\frac{80 + 60}{2} - 30)$
$\frac{60 - T}{10} = K (\frac{60 + T}{2} - 30)$
$T = 48^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366641 Two friends $A$ and $B$ are waiting for another friend for tea. $A$ took the tea in a cup and mixed the cold milk and then waits. $B$ took the tea in the cup and then mixed the cold milk when the friend comes. Then, the tea will be hotter in the cup of

1

A

2

B

3 tea will be equally hot in both cups

4 friend's cup

Explanation:

The rate of heat loss is proportional to the difference in temperature. The difference of temperature between the tea in cup $A$ and the surrounding is reduced, so it losses less heat, the tea in cup $B$ losses more heat because of large temperature difference.Hence, the tea in $cup A$ will be hotter.

PHXI11:THERMAL PROPERTIES OF MATTER

366642 The room temperature is $+ 20^{\circ} C$ when outside temperature is $- 20^{\circ} C$ and room temperature is $+ 10^{\circ} C$ when outside temperature is $- 40^{\circ} C$ . Find the temperature of the heater heating the room.

1

30^{\circ} C

2

20^{\circ} C

3

10^{\circ} C

4

60^{\circ} C

Explanation:

Let $T$ is the temperature of the heater. The rate of change of temperature is
$\frac{d T}{d t} = - K (T - T_{0})$
Here heat flows from heater to the room and and then from room the atmosphere. So at thermal equilibrium $K_{1}$ and $K_{2}$ are corresponding constants related to heater to room and room to atmosphere
In case (1)
$K_{1} (T - T_{r_{1}}) = K_{2} (T_{r_{1}} - T_{out 1})$
and In case (2)
$K_{1} (T - T_{r_{2}}) = K_{2} (T_{r_{2}} - T_{out 2})$
Dividing these equations
$\frac{T - T_{r_{1}}}{T - T_{r_{2}}} = \frac{T_{r_{1}} - T_{o u t 1}}{T_{r_{2}} - T_{o u t 1}}$
$= \frac{T - 20}{T - 10} = \frac{20 - (- 20)}{10 - (- 40)}$
$\Rightarrow T = 60^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366643 A system $S$ receives heat continuously from an electrical heater of power $10 W$ . The temperature of $S$ becomes constant at $50^{\circ} C$ when the surrounding temperature is $20^{\circ} C$ . After the heater is switched off, $S$ cools from ${35.1}^{\circ} C$ to ${34.9}^{\circ} C$ in 1 minute. The heat capacity of $S$ is

1

100 J /^{\circ} C

2

300 J /^{\circ} C

3

750 J /^{\circ} C

4

1500 J /^{\circ} C

Explanation:

At equilibrium the heat gained from heater is equal to the heat lost to the surrounds.
When the heater is on.
$\frac{d Q}{d t} = k (50 - 20) = 10$ ,
Where $k =$ constant
$\Rightarrow k = \frac{1}{3}$
At an average temperature of $35^{\circ} C$ , heat lost to the surroundings is
$\frac{d Q}{d t} = \frac{1}{3} (35 - 20) J / s = 5 J / s$
Heat lost in 1 minute
$= \frac{d Q}{d t} \times 60 J$
$= 5 \times 60 J = 300 J$
Decreases in temperature $= {0.2}^{\circ} C = Δ θ$
$Q = c Δ θ$
Heat capacity $= c = \frac{Q}{Δ θ} = \frac{300 J}{{0.2}^{0} C} = 1500 J /^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366639 A body cools from $80^{\circ} C$ to $50^{\circ} C$ in $5 min$ . Calculate the time it takes to cool from $60^{\circ} C$ to $30^{\circ} C$ . The temperature of the surroundings is $20^{\circ} C$ .

1

9 \min

2

7 \min

3

8 \min

4

10 \min

Explanation:

Given, initial temperature $T_{1} = 80^{\circ} C$
Final temperature $T_{2} = 50^{\circ} C$
Temperature of the surroundings $T_{0} = 20^{\circ} C$
$t_{1} = 5 min$
According to Newton's law of cooling.
$\begin{aligned} Rate of cooling, \frac{d T}{d t} = k [\frac{T_{1} + T_{2}}{2} - T_{0}] \\ \frac{(80 - 50)}{5} = k (\frac{80 + 50}{2} - 20) \\ \frac{30}{5} = k (65 - 20) \\ \Rightarrow k = \frac{6}{45} = \frac{2}{15} \end{aligned}$
In second case,
Initial temperature $T_{1}^{'} = 60^{\circ} C$
Final temperature $T_{2}^{'} = 30^{\circ} C$
$t^{'} = ?$
$N o w, \frac{(60 - 30)}{t^{'}} = \frac{2}{15} (\frac{60 + 30}{2} - 20)$
$\Rightarrow t^{'} = \frac{30 \times 15}{2 \times 25} = 9 min .$

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366640 A body cools from $80^{\circ} C$ to $60^{\circ} C$ in $10 \min$ , when the temperature of surroundings is $30^{\circ} C$ . The temperature of the body after next $10 min$ will be

1

40^{\circ} C

2

43^{\circ} C

3

48^{\circ} C

4

52^{\circ} C

Explanation:

From Newton's law of cooling
$\frac{(80 - 60)}{10} = K (\frac{80 + 60}{2} - 30)$
$\frac{60 - T}{10} = K (\frac{60 + T}{2} - 30)$
$T = 48^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366641 Two friends $A$ and $B$ are waiting for another friend for tea. $A$ took the tea in a cup and mixed the cold milk and then waits. $B$ took the tea in the cup and then mixed the cold milk when the friend comes. Then, the tea will be hotter in the cup of

1

A

2

B

3 tea will be equally hot in both cups

4 friend's cup

Explanation:

The rate of heat loss is proportional to the difference in temperature. The difference of temperature between the tea in cup $A$ and the surrounding is reduced, so it losses less heat, the tea in cup $B$ losses more heat because of large temperature difference.Hence, the tea in $cup A$ will be hotter.

PHXI11:THERMAL PROPERTIES OF MATTER

366642 The room temperature is $+ 20^{\circ} C$ when outside temperature is $- 20^{\circ} C$ and room temperature is $+ 10^{\circ} C$ when outside temperature is $- 40^{\circ} C$ . Find the temperature of the heater heating the room.

1

30^{\circ} C

2

20^{\circ} C

3

10^{\circ} C

4

60^{\circ} C

Explanation:

Let $T$ is the temperature of the heater. The rate of change of temperature is
$\frac{d T}{d t} = - K (T - T_{0})$
Here heat flows from heater to the room and and then from room the atmosphere. So at thermal equilibrium $K_{1}$ and $K_{2}$ are corresponding constants related to heater to room and room to atmosphere
In case (1)
$K_{1} (T - T_{r_{1}}) = K_{2} (T_{r_{1}} - T_{out 1})$
and In case (2)
$K_{1} (T - T_{r_{2}}) = K_{2} (T_{r_{2}} - T_{out 2})$
Dividing these equations
$\frac{T - T_{r_{1}}}{T - T_{r_{2}}} = \frac{T_{r_{1}} - T_{o u t 1}}{T_{r_{2}} - T_{o u t 1}}$
$= \frac{T - 20}{T - 10} = \frac{20 - (- 20)}{10 - (- 40)}$
$\Rightarrow T = 60^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366643 A system $S$ receives heat continuously from an electrical heater of power $10 W$ . The temperature of $S$ becomes constant at $50^{\circ} C$ when the surrounding temperature is $20^{\circ} C$ . After the heater is switched off, $S$ cools from ${35.1}^{\circ} C$ to ${34.9}^{\circ} C$ in 1 minute. The heat capacity of $S$ is

1

100 J /^{\circ} C

2

300 J /^{\circ} C

3

750 J /^{\circ} C

4

1500 J /^{\circ} C

Explanation:

At equilibrium the heat gained from heater is equal to the heat lost to the surrounds.
When the heater is on.
$\frac{d Q}{d t} = k (50 - 20) = 10$ ,
Where $k =$ constant
$\Rightarrow k = \frac{1}{3}$
At an average temperature of $35^{\circ} C$ , heat lost to the surroundings is
$\frac{d Q}{d t} = \frac{1}{3} (35 - 20) J / s = 5 J / s$
Heat lost in 1 minute
$= \frac{d Q}{d t} \times 60 J$
$= 5 \times 60 J = 300 J$
Decreases in temperature $= {0.2}^{\circ} C = Δ θ$
$Q = c Δ θ$
Heat capacity $= c = \frac{Q}{Δ θ} = \frac{300 J}{{0.2}^{0} C} = 1500 J /^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366639 A body cools from $80^{\circ} C$ to $50^{\circ} C$ in $5 min$ . Calculate the time it takes to cool from $60^{\circ} C$ to $30^{\circ} C$ . The temperature of the surroundings is $20^{\circ} C$ .

1

9 \min

2

7 \min

3

8 \min

4

10 \min

Explanation:

Given, initial temperature $T_{1} = 80^{\circ} C$
Final temperature $T_{2} = 50^{\circ} C$
Temperature of the surroundings $T_{0} = 20^{\circ} C$
$t_{1} = 5 min$
According to Newton's law of cooling.
$\begin{aligned} Rate of cooling, \frac{d T}{d t} = k [\frac{T_{1} + T_{2}}{2} - T_{0}] \\ \frac{(80 - 50)}{5} = k (\frac{80 + 50}{2} - 20) \\ \frac{30}{5} = k (65 - 20) \\ \Rightarrow k = \frac{6}{45} = \frac{2}{15} \end{aligned}$
In second case,
Initial temperature $T_{1}^{'} = 60^{\circ} C$
Final temperature $T_{2}^{'} = 30^{\circ} C$
$t^{'} = ?$
$N o w, \frac{(60 - 30)}{t^{'}} = \frac{2}{15} (\frac{60 + 30}{2} - 20)$
$\Rightarrow t^{'} = \frac{30 \times 15}{2 \times 25} = 9 min .$

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366640 A body cools from $80^{\circ} C$ to $60^{\circ} C$ in $10 \min$ , when the temperature of surroundings is $30^{\circ} C$ . The temperature of the body after next $10 min$ will be

1

40^{\circ} C

2

43^{\circ} C

3

48^{\circ} C

4

52^{\circ} C

Explanation:

From Newton's law of cooling
$\frac{(80 - 60)}{10} = K (\frac{80 + 60}{2} - 30)$
$\frac{60 - T}{10} = K (\frac{60 + T}{2} - 30)$
$T = 48^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366641 Two friends $A$ and $B$ are waiting for another friend for tea. $A$ took the tea in a cup and mixed the cold milk and then waits. $B$ took the tea in the cup and then mixed the cold milk when the friend comes. Then, the tea will be hotter in the cup of

1

A

2

B

3 tea will be equally hot in both cups

4 friend's cup

Explanation:

The rate of heat loss is proportional to the difference in temperature. The difference of temperature between the tea in cup $A$ and the surrounding is reduced, so it losses less heat, the tea in cup $B$ losses more heat because of large temperature difference.Hence, the tea in $cup A$ will be hotter.

PHXI11:THERMAL PROPERTIES OF MATTER

366642 The room temperature is $+ 20^{\circ} C$ when outside temperature is $- 20^{\circ} C$ and room temperature is $+ 10^{\circ} C$ when outside temperature is $- 40^{\circ} C$ . Find the temperature of the heater heating the room.

1

30^{\circ} C

2

20^{\circ} C

3

10^{\circ} C

4

60^{\circ} C

Explanation:

Let $T$ is the temperature of the heater. The rate of change of temperature is
$\frac{d T}{d t} = - K (T - T_{0})$
Here heat flows from heater to the room and and then from room the atmosphere. So at thermal equilibrium $K_{1}$ and $K_{2}$ are corresponding constants related to heater to room and room to atmosphere
In case (1)
$K_{1} (T - T_{r_{1}}) = K_{2} (T_{r_{1}} - T_{out 1})$
and In case (2)
$K_{1} (T - T_{r_{2}}) = K_{2} (T_{r_{2}} - T_{out 2})$
Dividing these equations
$\frac{T - T_{r_{1}}}{T - T_{r_{2}}} = \frac{T_{r_{1}} - T_{o u t 1}}{T_{r_{2}} - T_{o u t 1}}$
$= \frac{T - 20}{T - 10} = \frac{20 - (- 20)}{10 - (- 40)}$
$\Rightarrow T = 60^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366643 A system $S$ receives heat continuously from an electrical heater of power $10 W$ . The temperature of $S$ becomes constant at $50^{\circ} C$ when the surrounding temperature is $20^{\circ} C$ . After the heater is switched off, $S$ cools from ${35.1}^{\circ} C$ to ${34.9}^{\circ} C$ in 1 minute. The heat capacity of $S$ is

1

100 J /^{\circ} C

2

300 J /^{\circ} C

3

750 J /^{\circ} C

4

1500 J /^{\circ} C

Explanation:

At equilibrium the heat gained from heater is equal to the heat lost to the surrounds.
When the heater is on.
$\frac{d Q}{d t} = k (50 - 20) = 10$ ,
Where $k =$ constant
$\Rightarrow k = \frac{1}{3}$
At an average temperature of $35^{\circ} C$ , heat lost to the surroundings is
$\frac{d Q}{d t} = \frac{1}{3} (35 - 20) J / s = 5 J / s$
Heat lost in 1 minute
$= \frac{d Q}{d t} \times 60 J$
$= 5 \times 60 J = 300 J$
Decreases in temperature $= {0.2}^{\circ} C = Δ θ$
$Q = c Δ θ$
Heat capacity $= c = \frac{Q}{Δ θ} = \frac{300 J}{{0.2}^{0} C} = 1500 J /^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366639 A body cools from $80^{\circ} C$ to $50^{\circ} C$ in $5 min$ . Calculate the time it takes to cool from $60^{\circ} C$ to $30^{\circ} C$ . The temperature of the surroundings is $20^{\circ} C$ .

1

9 \min

2

7 \min

3

8 \min

4

10 \min

Explanation:

Given, initial temperature $T_{1} = 80^{\circ} C$
Final temperature $T_{2} = 50^{\circ} C$
Temperature of the surroundings $T_{0} = 20^{\circ} C$
$t_{1} = 5 min$
According to Newton's law of cooling.
$\begin{aligned} Rate of cooling, \frac{d T}{d t} = k [\frac{T_{1} + T_{2}}{2} - T_{0}] \\ \frac{(80 - 50)}{5} = k (\frac{80 + 50}{2} - 20) \\ \frac{30}{5} = k (65 - 20) \\ \Rightarrow k = \frac{6}{45} = \frac{2}{15} \end{aligned}$
In second case,
Initial temperature $T_{1}^{'} = 60^{\circ} C$
Final temperature $T_{2}^{'} = 30^{\circ} C$
$t^{'} = ?$
$N o w, \frac{(60 - 30)}{t^{'}} = \frac{2}{15} (\frac{60 + 30}{2} - 20)$
$\Rightarrow t^{'} = \frac{30 \times 15}{2 \times 25} = 9 min .$

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366640 A body cools from $80^{\circ} C$ to $60^{\circ} C$ in $10 \min$ , when the temperature of surroundings is $30^{\circ} C$ . The temperature of the body after next $10 min$ will be

1

40^{\circ} C

2

43^{\circ} C

3

48^{\circ} C

4

52^{\circ} C

Explanation:

From Newton's law of cooling
$\frac{(80 - 60)}{10} = K (\frac{80 + 60}{2} - 30)$
$\frac{60 - T}{10} = K (\frac{60 + T}{2} - 30)$
$T = 48^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366641 Two friends $A$ and $B$ are waiting for another friend for tea. $A$ took the tea in a cup and mixed the cold milk and then waits. $B$ took the tea in the cup and then mixed the cold milk when the friend comes. Then, the tea will be hotter in the cup of

1

A

2

B

3 tea will be equally hot in both cups

4 friend's cup

Explanation:

The rate of heat loss is proportional to the difference in temperature. The difference of temperature between the tea in cup $A$ and the surrounding is reduced, so it losses less heat, the tea in cup $B$ losses more heat because of large temperature difference.Hence, the tea in $cup A$ will be hotter.

PHXI11:THERMAL PROPERTIES OF MATTER

366642 The room temperature is $+ 20^{\circ} C$ when outside temperature is $- 20^{\circ} C$ and room temperature is $+ 10^{\circ} C$ when outside temperature is $- 40^{\circ} C$ . Find the temperature of the heater heating the room.

1

30^{\circ} C

2

20^{\circ} C

3

10^{\circ} C

4

60^{\circ} C

Explanation:

Let $T$ is the temperature of the heater. The rate of change of temperature is
$\frac{d T}{d t} = - K (T - T_{0})$
Here heat flows from heater to the room and and then from room the atmosphere. So at thermal equilibrium $K_{1}$ and $K_{2}$ are corresponding constants related to heater to room and room to atmosphere
In case (1)
$K_{1} (T - T_{r_{1}}) = K_{2} (T_{r_{1}} - T_{out 1})$
and In case (2)
$K_{1} (T - T_{r_{2}}) = K_{2} (T_{r_{2}} - T_{out 2})$
Dividing these equations
$\frac{T - T_{r_{1}}}{T - T_{r_{2}}} = \frac{T_{r_{1}} - T_{o u t 1}}{T_{r_{2}} - T_{o u t 1}}$
$= \frac{T - 20}{T - 10} = \frac{20 - (- 20)}{10 - (- 40)}$
$\Rightarrow T = 60^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366643 A system $S$ receives heat continuously from an electrical heater of power $10 W$ . The temperature of $S$ becomes constant at $50^{\circ} C$ when the surrounding temperature is $20^{\circ} C$ . After the heater is switched off, $S$ cools from ${35.1}^{\circ} C$ to ${34.9}^{\circ} C$ in 1 minute. The heat capacity of $S$ is

1

100 J /^{\circ} C

2

300 J /^{\circ} C

3

750 J /^{\circ} C

4

1500 J /^{\circ} C

Explanation:

At equilibrium the heat gained from heater is equal to the heat lost to the surrounds.
When the heater is on.
$\frac{d Q}{d t} = k (50 - 20) = 10$ ,
Where $k =$ constant
$\Rightarrow k = \frac{1}{3}$
At an average temperature of $35^{\circ} C$ , heat lost to the surroundings is
$\frac{d Q}{d t} = \frac{1}{3} (35 - 20) J / s = 5 J / s$
Heat lost in 1 minute
$= \frac{d Q}{d t} \times 60 J$
$= 5 \times 60 J = 300 J$
Decreases in temperature $= {0.2}^{\circ} C = Δ θ$
$Q = c Δ θ$
Heat capacity $= c = \frac{Q}{Δ θ} = \frac{300 J}{{0.2}^{0} C} = 1500 J /^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366639 A body cools from $80^{\circ} C$ to $50^{\circ} C$ in $5 min$ . Calculate the time it takes to cool from $60^{\circ} C$ to $30^{\circ} C$ . The temperature of the surroundings is $20^{\circ} C$ .

1

9 \min

2

7 \min

3

8 \min

4

10 \min

Explanation:

Given, initial temperature $T_{1} = 80^{\circ} C$
Final temperature $T_{2} = 50^{\circ} C$
Temperature of the surroundings $T_{0} = 20^{\circ} C$
$t_{1} = 5 min$
According to Newton's law of cooling.
$\begin{aligned} Rate of cooling, \frac{d T}{d t} = k [\frac{T_{1} + T_{2}}{2} - T_{0}] \\ \frac{(80 - 50)}{5} = k (\frac{80 + 50}{2} - 20) \\ \frac{30}{5} = k (65 - 20) \\ \Rightarrow k = \frac{6}{45} = \frac{2}{15} \end{aligned}$
In second case,
Initial temperature $T_{1}^{'} = 60^{\circ} C$
Final temperature $T_{2}^{'} = 30^{\circ} C$
$t^{'} = ?$
$N o w, \frac{(60 - 30)}{t^{'}} = \frac{2}{15} (\frac{60 + 30}{2} - 20)$
$\Rightarrow t^{'} = \frac{30 \times 15}{2 \times 25} = 9 min .$

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366640 A body cools from $80^{\circ} C$ to $60^{\circ} C$ in $10 \min$ , when the temperature of surroundings is $30^{\circ} C$ . The temperature of the body after next $10 min$ will be

1

40^{\circ} C

2

43^{\circ} C

3

48^{\circ} C

4

52^{\circ} C

Explanation:

From Newton's law of cooling
$\frac{(80 - 60)}{10} = K (\frac{80 + 60}{2} - 30)$
$\frac{60 - T}{10} = K (\frac{60 + T}{2} - 30)$
$T = 48^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366641 Two friends $A$ and $B$ are waiting for another friend for tea. $A$ took the tea in a cup and mixed the cold milk and then waits. $B$ took the tea in the cup and then mixed the cold milk when the friend comes. Then, the tea will be hotter in the cup of

1

A

2

B

3 tea will be equally hot in both cups

4 friend's cup

Explanation:

The rate of heat loss is proportional to the difference in temperature. The difference of temperature between the tea in cup $A$ and the surrounding is reduced, so it losses less heat, the tea in cup $B$ losses more heat because of large temperature difference.Hence, the tea in $cup A$ will be hotter.

PHXI11:THERMAL PROPERTIES OF MATTER

366642 The room temperature is $+ 20^{\circ} C$ when outside temperature is $- 20^{\circ} C$ and room temperature is $+ 10^{\circ} C$ when outside temperature is $- 40^{\circ} C$ . Find the temperature of the heater heating the room.

1

30^{\circ} C

2

20^{\circ} C

3

10^{\circ} C

4

60^{\circ} C

Explanation:

Let $T$ is the temperature of the heater. The rate of change of temperature is
$\frac{d T}{d t} = - K (T - T_{0})$
Here heat flows from heater to the room and and then from room the atmosphere. So at thermal equilibrium $K_{1}$ and $K_{2}$ are corresponding constants related to heater to room and room to atmosphere
In case (1)
$K_{1} (T - T_{r_{1}}) = K_{2} (T_{r_{1}} - T_{out 1})$
and In case (2)
$K_{1} (T - T_{r_{2}}) = K_{2} (T_{r_{2}} - T_{out 2})$
Dividing these equations
$\frac{T - T_{r_{1}}}{T - T_{r_{2}}} = \frac{T_{r_{1}} - T_{o u t 1}}{T_{r_{2}} - T_{o u t 1}}$
$= \frac{T - 20}{T - 10} = \frac{20 - (- 20)}{10 - (- 40)}$
$\Rightarrow T = 60^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366643 A system $S$ receives heat continuously from an electrical heater of power $10 W$ . The temperature of $S$ becomes constant at $50^{\circ} C$ when the surrounding temperature is $20^{\circ} C$ . After the heater is switched off, $S$ cools from ${35.1}^{\circ} C$ to ${34.9}^{\circ} C$ in 1 minute. The heat capacity of $S$ is

1

100 J /^{\circ} C

2

300 J /^{\circ} C

3

750 J /^{\circ} C

4

1500 J /^{\circ} C

Explanation:

At equilibrium the heat gained from heater is equal to the heat lost to the surrounds.
When the heater is on.
$\frac{d Q}{d t} = k (50 - 20) = 10$ ,
Where $k =$ constant
$\Rightarrow k = \frac{1}{3}$
At an average temperature of $35^{\circ} C$ , heat lost to the surroundings is
$\frac{d Q}{d t} = \frac{1}{3} (35 - 20) J / s = 5 J / s$
Heat lost in 1 minute
$= \frac{d Q}{d t} \times 60 J$
$= 5 \times 60 J = 300 J$
Decreases in temperature $= {0.2}^{\circ} C = Δ θ$
$Q = c Δ θ$
Heat capacity $= c = \frac{Q}{Δ θ} = \frac{300 J}{{0.2}^{0} C} = 1500 J /^{\circ} C$

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