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PHXI11:THERMAL PROPERTIES OF MATTER

366311 Three liquids with masses $m_{1}, m_{2}, m_{3}$ are thoroughly mixed. If their specific heats are $c_{1}, c_{2}, c_{3}$ and their temperatures $T_{1}, T_{2}, T_{3}$ respectively, then the temperature of the mixture is

1

\frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}}

2

\frac{c_{1} T_{1} + c_{2} T_{2} + c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}}

3

\frac{m_{1} T_{1} + m_{2} T_{2} + m_{3} T_{3}}{c_{1} T_{1} + c_{2} T_{2} + c_{3} T_{3}}

4

\frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} T_{1} + m_{2} T_{2} + m_{3} T_{3}}

Explanation:

Let the final temperature be $T^{\circ} C$ .
The net heat change of the system is zero.
$\begin{matrix} Δ q_{1} + Δ q_{2} + Δ q_{3} = 0 \\ m_{1} c_{1} (T - T_{1}) + m_{2} c_{2} (T - T_{2}) + m_{3} c_{3} (T - T_{3}) = 0 \\ \Rightarrow T = \frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}} \end{matrix}$

PHXI11:THERMAL PROPERTIES OF MATTER

366312 An electrical heater having coil is immersed in a calorimeter containing $360 g$ of water at $90^{\circ} C$ . The coil consumes energy at the rate of $90 W$ . The water equivalent of calorimeter and coil is $40 g$ . The temperature of water after $10 min$ is

1

30^{\circ} C

2

{4.214}^{\circ} C

3

{42.14}^{\circ} C

4 None of these

Explanation:

Energy supplied by the heater to the system in $10 min$
$Q_{1} = P \times t = 90 \times 10 \times 60 = 54000 J$
Now if $θ$ is the final temperature of the system, energy absorbed by it to change its temperature from $10^{\circ} C$ to $θ^{\circ} C$ is
$Q_{2} = (m s Δ T)_{w a t e r} + (m s Δ T)_{c o i l + c a l o r i m e t e r}$
$= [360 \times 1 (θ - 10) + 40 (θ - 10)] 4200$
$= 400 (θ - 10) 4200$
$Q_{1} = Q_{2} \Rightarrow θ = 10^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366313 In defining the specific heat, temperature is represented in $^{\circ} F$ instead of $^{\circ} C$ . Then the value of specific heat will

1 Decrease

2 Increase

3 Remain constant

4 Be converted to heat capacity

Explanation:

The coversion factor between $^{\circ} F$ and $^{\circ} C$ is
$\begin{aligned} ^{\circ} F =^{\circ} C (\frac{9}{5}) + 32 \\ \Rightarrow Δ^{\circ} F = \frac{9}{5} (Δ^{\circ} C) \end{aligned}$
As $Δ^{\circ} F \neq Δ^{\circ} C$ the heat capacity is not same in both scales.
since $s = \frac{1}{m} \frac{Δ Q}{Δ T}$
As $Δ Q$ and $m$ are same $s Δ T =$ constant
$S_{c} Δ T_{c} = S_{F} Δ T_{F}$
As $Δ T_{F} > Δ T_{c} \Rightarrow S_{F} < S_{c}$

PHXI11:THERMAL PROPERTIES OF MATTER

366314 The heat capacity of a material depends upon

1 The change in temperature

2 Mass

3 Nature of substance

4 All of the above

PHXI11:THERMAL PROPERTIES OF MATTER

366315 During illness an $80 k g$ man ran a fever of ${102.2}^{\circ} F$ instead of normal body temperature of ${98.6}^{\circ} F$ . Assuming that human body is mostly water, how much heat is required to raise his temperature by that amount?

1

100 k c a l

2

160 k c a l

3

50 k c a l

4

92 k c a l

Explanation:

Since ${102.2}^{\circ} F \to 39^{\circ} C a n d {98.6}^{\circ} F \to 37^{\circ} C$
Hence, $Δ Q = m s Δ T = 80 \times 1000 \times (39 - 37)$
$16 \times 10^{4} c a l = 160 k c a l$

PHXI11:THERMAL PROPERTIES OF MATTER

366311 Three liquids with masses $m_{1}, m_{2}, m_{3}$ are thoroughly mixed. If their specific heats are $c_{1}, c_{2}, c_{3}$ and their temperatures $T_{1}, T_{2}, T_{3}$ respectively, then the temperature of the mixture is

1

\frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}}

2

\frac{c_{1} T_{1} + c_{2} T_{2} + c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}}

3

\frac{m_{1} T_{1} + m_{2} T_{2} + m_{3} T_{3}}{c_{1} T_{1} + c_{2} T_{2} + c_{3} T_{3}}

4

\frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} T_{1} + m_{2} T_{2} + m_{3} T_{3}}

Explanation:

Let the final temperature be $T^{\circ} C$ .
The net heat change of the system is zero.
$\begin{matrix} Δ q_{1} + Δ q_{2} + Δ q_{3} = 0 \\ m_{1} c_{1} (T - T_{1}) + m_{2} c_{2} (T - T_{2}) + m_{3} c_{3} (T - T_{3}) = 0 \\ \Rightarrow T = \frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}} \end{matrix}$

PHXI11:THERMAL PROPERTIES OF MATTER

366312 An electrical heater having coil is immersed in a calorimeter containing $360 g$ of water at $90^{\circ} C$ . The coil consumes energy at the rate of $90 W$ . The water equivalent of calorimeter and coil is $40 g$ . The temperature of water after $10 min$ is

1

30^{\circ} C

2

{4.214}^{\circ} C

3

{42.14}^{\circ} C

4 None of these

Explanation:

Energy supplied by the heater to the system in $10 min$
$Q_{1} = P \times t = 90 \times 10 \times 60 = 54000 J$
Now if $θ$ is the final temperature of the system, energy absorbed by it to change its temperature from $10^{\circ} C$ to $θ^{\circ} C$ is
$Q_{2} = (m s Δ T)_{w a t e r} + (m s Δ T)_{c o i l + c a l o r i m e t e r}$
$= [360 \times 1 (θ - 10) + 40 (θ - 10)] 4200$
$= 400 (θ - 10) 4200$
$Q_{1} = Q_{2} \Rightarrow θ = 10^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366313 In defining the specific heat, temperature is represented in $^{\circ} F$ instead of $^{\circ} C$ . Then the value of specific heat will

1 Decrease

2 Increase

3 Remain constant

4 Be converted to heat capacity

Explanation:

The coversion factor between $^{\circ} F$ and $^{\circ} C$ is
$\begin{aligned} ^{\circ} F =^{\circ} C (\frac{9}{5}) + 32 \\ \Rightarrow Δ^{\circ} F = \frac{9}{5} (Δ^{\circ} C) \end{aligned}$
As $Δ^{\circ} F \neq Δ^{\circ} C$ the heat capacity is not same in both scales.
since $s = \frac{1}{m} \frac{Δ Q}{Δ T}$
As $Δ Q$ and $m$ are same $s Δ T =$ constant
$S_{c} Δ T_{c} = S_{F} Δ T_{F}$
As $Δ T_{F} > Δ T_{c} \Rightarrow S_{F} < S_{c}$

PHXI11:THERMAL PROPERTIES OF MATTER

366314 The heat capacity of a material depends upon

1 The change in temperature

2 Mass

3 Nature of substance

4 All of the above

PHXI11:THERMAL PROPERTIES OF MATTER

366315 During illness an $80 k g$ man ran a fever of ${102.2}^{\circ} F$ instead of normal body temperature of ${98.6}^{\circ} F$ . Assuming that human body is mostly water, how much heat is required to raise his temperature by that amount?

1

100 k c a l

2

160 k c a l

3

50 k c a l

4

92 k c a l

Explanation:

Since ${102.2}^{\circ} F \to 39^{\circ} C a n d {98.6}^{\circ} F \to 37^{\circ} C$
Hence, $Δ Q = m s Δ T = 80 \times 1000 \times (39 - 37)$
$16 \times 10^{4} c a l = 160 k c a l$

PHXI11:THERMAL PROPERTIES OF MATTER

366311 Three liquids with masses $m_{1}, m_{2}, m_{3}$ are thoroughly mixed. If their specific heats are $c_{1}, c_{2}, c_{3}$ and their temperatures $T_{1}, T_{2}, T_{3}$ respectively, then the temperature of the mixture is

1

\frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}}

2

\frac{c_{1} T_{1} + c_{2} T_{2} + c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}}

3

\frac{m_{1} T_{1} + m_{2} T_{2} + m_{3} T_{3}}{c_{1} T_{1} + c_{2} T_{2} + c_{3} T_{3}}

4

\frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} T_{1} + m_{2} T_{2} + m_{3} T_{3}}

Explanation:

Let the final temperature be $T^{\circ} C$ .
The net heat change of the system is zero.
$\begin{matrix} Δ q_{1} + Δ q_{2} + Δ q_{3} = 0 \\ m_{1} c_{1} (T - T_{1}) + m_{2} c_{2} (T - T_{2}) + m_{3} c_{3} (T - T_{3}) = 0 \\ \Rightarrow T = \frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}} \end{matrix}$

PHXI11:THERMAL PROPERTIES OF MATTER

366312 An electrical heater having coil is immersed in a calorimeter containing $360 g$ of water at $90^{\circ} C$ . The coil consumes energy at the rate of $90 W$ . The water equivalent of calorimeter and coil is $40 g$ . The temperature of water after $10 min$ is

1

30^{\circ} C

2

{4.214}^{\circ} C

3

{42.14}^{\circ} C

4 None of these

Explanation:

Energy supplied by the heater to the system in $10 min$
$Q_{1} = P \times t = 90 \times 10 \times 60 = 54000 J$
Now if $θ$ is the final temperature of the system, energy absorbed by it to change its temperature from $10^{\circ} C$ to $θ^{\circ} C$ is
$Q_{2} = (m s Δ T)_{w a t e r} + (m s Δ T)_{c o i l + c a l o r i m e t e r}$
$= [360 \times 1 (θ - 10) + 40 (θ - 10)] 4200$
$= 400 (θ - 10) 4200$
$Q_{1} = Q_{2} \Rightarrow θ = 10^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366313 In defining the specific heat, temperature is represented in $^{\circ} F$ instead of $^{\circ} C$ . Then the value of specific heat will

1 Decrease

2 Increase

3 Remain constant

4 Be converted to heat capacity

Explanation:

The coversion factor between $^{\circ} F$ and $^{\circ} C$ is
$\begin{aligned} ^{\circ} F =^{\circ} C (\frac{9}{5}) + 32 \\ \Rightarrow Δ^{\circ} F = \frac{9}{5} (Δ^{\circ} C) \end{aligned}$
As $Δ^{\circ} F \neq Δ^{\circ} C$ the heat capacity is not same in both scales.
since $s = \frac{1}{m} \frac{Δ Q}{Δ T}$
As $Δ Q$ and $m$ are same $s Δ T =$ constant
$S_{c} Δ T_{c} = S_{F} Δ T_{F}$
As $Δ T_{F} > Δ T_{c} \Rightarrow S_{F} < S_{c}$

PHXI11:THERMAL PROPERTIES OF MATTER

366314 The heat capacity of a material depends upon

1 The change in temperature

2 Mass

3 Nature of substance

4 All of the above

PHXI11:THERMAL PROPERTIES OF MATTER

366315 During illness an $80 k g$ man ran a fever of ${102.2}^{\circ} F$ instead of normal body temperature of ${98.6}^{\circ} F$ . Assuming that human body is mostly water, how much heat is required to raise his temperature by that amount?

1

100 k c a l

2

160 k c a l

3

50 k c a l

4

92 k c a l

Explanation:

Since ${102.2}^{\circ} F \to 39^{\circ} C a n d {98.6}^{\circ} F \to 37^{\circ} C$
Hence, $Δ Q = m s Δ T = 80 \times 1000 \times (39 - 37)$
$16 \times 10^{4} c a l = 160 k c a l$

PHXI11:THERMAL PROPERTIES OF MATTER

366311 Three liquids with masses $m_{1}, m_{2}, m_{3}$ are thoroughly mixed. If their specific heats are $c_{1}, c_{2}, c_{3}$ and their temperatures $T_{1}, T_{2}, T_{3}$ respectively, then the temperature of the mixture is

1

\frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}}

2

\frac{c_{1} T_{1} + c_{2} T_{2} + c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}}

3

\frac{m_{1} T_{1} + m_{2} T_{2} + m_{3} T_{3}}{c_{1} T_{1} + c_{2} T_{2} + c_{3} T_{3}}

4

\frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} T_{1} + m_{2} T_{2} + m_{3} T_{3}}

Explanation:

Let the final temperature be $T^{\circ} C$ .
The net heat change of the system is zero.
$\begin{matrix} Δ q_{1} + Δ q_{2} + Δ q_{3} = 0 \\ m_{1} c_{1} (T - T_{1}) + m_{2} c_{2} (T - T_{2}) + m_{3} c_{3} (T - T_{3}) = 0 \\ \Rightarrow T = \frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}} \end{matrix}$

PHXI11:THERMAL PROPERTIES OF MATTER

366312 An electrical heater having coil is immersed in a calorimeter containing $360 g$ of water at $90^{\circ} C$ . The coil consumes energy at the rate of $90 W$ . The water equivalent of calorimeter and coil is $40 g$ . The temperature of water after $10 min$ is

1

30^{\circ} C

2

{4.214}^{\circ} C

3

{42.14}^{\circ} C

4 None of these

Explanation:

Energy supplied by the heater to the system in $10 min$
$Q_{1} = P \times t = 90 \times 10 \times 60 = 54000 J$
Now if $θ$ is the final temperature of the system, energy absorbed by it to change its temperature from $10^{\circ} C$ to $θ^{\circ} C$ is
$Q_{2} = (m s Δ T)_{w a t e r} + (m s Δ T)_{c o i l + c a l o r i m e t e r}$
$= [360 \times 1 (θ - 10) + 40 (θ - 10)] 4200$
$= 400 (θ - 10) 4200$
$Q_{1} = Q_{2} \Rightarrow θ = 10^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366313 In defining the specific heat, temperature is represented in $^{\circ} F$ instead of $^{\circ} C$ . Then the value of specific heat will

1 Decrease

2 Increase

3 Remain constant

4 Be converted to heat capacity

Explanation:

The coversion factor between $^{\circ} F$ and $^{\circ} C$ is
$\begin{aligned} ^{\circ} F =^{\circ} C (\frac{9}{5}) + 32 \\ \Rightarrow Δ^{\circ} F = \frac{9}{5} (Δ^{\circ} C) \end{aligned}$
As $Δ^{\circ} F \neq Δ^{\circ} C$ the heat capacity is not same in both scales.
since $s = \frac{1}{m} \frac{Δ Q}{Δ T}$
As $Δ Q$ and $m$ are same $s Δ T =$ constant
$S_{c} Δ T_{c} = S_{F} Δ T_{F}$
As $Δ T_{F} > Δ T_{c} \Rightarrow S_{F} < S_{c}$

PHXI11:THERMAL PROPERTIES OF MATTER

366314 The heat capacity of a material depends upon

1 The change in temperature

2 Mass

3 Nature of substance

4 All of the above

PHXI11:THERMAL PROPERTIES OF MATTER

366315 During illness an $80 k g$ man ran a fever of ${102.2}^{\circ} F$ instead of normal body temperature of ${98.6}^{\circ} F$ . Assuming that human body is mostly water, how much heat is required to raise his temperature by that amount?

1

100 k c a l

2

160 k c a l

3

50 k c a l

4

92 k c a l

Explanation:

Since ${102.2}^{\circ} F \to 39^{\circ} C a n d {98.6}^{\circ} F \to 37^{\circ} C$
Hence, $Δ Q = m s Δ T = 80 \times 1000 \times (39 - 37)$
$16 \times 10^{4} c a l = 160 k c a l$

PHXI11:THERMAL PROPERTIES OF MATTER

366311 Three liquids with masses $m_{1}, m_{2}, m_{3}$ are thoroughly mixed. If their specific heats are $c_{1}, c_{2}, c_{3}$ and their temperatures $T_{1}, T_{2}, T_{3}$ respectively, then the temperature of the mixture is

1

\frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}}

2

\frac{c_{1} T_{1} + c_{2} T_{2} + c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}}

3

\frac{m_{1} T_{1} + m_{2} T_{2} + m_{3} T_{3}}{c_{1} T_{1} + c_{2} T_{2} + c_{3} T_{3}}

4

\frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} T_{1} + m_{2} T_{2} + m_{3} T_{3}}

Explanation:

Let the final temperature be $T^{\circ} C$ .
The net heat change of the system is zero.
$\begin{matrix} Δ q_{1} + Δ q_{2} + Δ q_{3} = 0 \\ m_{1} c_{1} (T - T_{1}) + m_{2} c_{2} (T - T_{2}) + m_{3} c_{3} (T - T_{3}) = 0 \\ \Rightarrow T = \frac{m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} + m_{3} c_{3} T_{3}}{m_{1} c_{1} + m_{2} c_{2} + m_{3} c_{3}} \end{matrix}$

PHXI11:THERMAL PROPERTIES OF MATTER

366312 An electrical heater having coil is immersed in a calorimeter containing $360 g$ of water at $90^{\circ} C$ . The coil consumes energy at the rate of $90 W$ . The water equivalent of calorimeter and coil is $40 g$ . The temperature of water after $10 min$ is

1

30^{\circ} C

2

{4.214}^{\circ} C

3

{42.14}^{\circ} C

4 None of these

Explanation:

Energy supplied by the heater to the system in $10 min$
$Q_{1} = P \times t = 90 \times 10 \times 60 = 54000 J$
Now if $θ$ is the final temperature of the system, energy absorbed by it to change its temperature from $10^{\circ} C$ to $θ^{\circ} C$ is
$Q_{2} = (m s Δ T)_{w a t e r} + (m s Δ T)_{c o i l + c a l o r i m e t e r}$
$= [360 \times 1 (θ - 10) + 40 (θ - 10)] 4200$
$= 400 (θ - 10) 4200$
$Q_{1} = Q_{2} \Rightarrow θ = 10^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366313 In defining the specific heat, temperature is represented in $^{\circ} F$ instead of $^{\circ} C$ . Then the value of specific heat will

1 Decrease

2 Increase

3 Remain constant

4 Be converted to heat capacity

Explanation:

The coversion factor between $^{\circ} F$ and $^{\circ} C$ is
$\begin{aligned} ^{\circ} F =^{\circ} C (\frac{9}{5}) + 32 \\ \Rightarrow Δ^{\circ} F = \frac{9}{5} (Δ^{\circ} C) \end{aligned}$
As $Δ^{\circ} F \neq Δ^{\circ} C$ the heat capacity is not same in both scales.
since $s = \frac{1}{m} \frac{Δ Q}{Δ T}$
As $Δ Q$ and $m$ are same $s Δ T =$ constant
$S_{c} Δ T_{c} = S_{F} Δ T_{F}$
As $Δ T_{F} > Δ T_{c} \Rightarrow S_{F} < S_{c}$

PHXI11:THERMAL PROPERTIES OF MATTER

366314 The heat capacity of a material depends upon

1 The change in temperature

2 Mass

3 Nature of substance

4 All of the above

PHXI11:THERMAL PROPERTIES OF MATTER

366315 During illness an $80 k g$ man ran a fever of ${102.2}^{\circ} F$ instead of normal body temperature of ${98.6}^{\circ} F$ . Assuming that human body is mostly water, how much heat is required to raise his temperature by that amount?

1

100 k c a l

2

160 k c a l

3

50 k c a l

4

92 k c a l

Explanation:

Since ${102.2}^{\circ} F \to 39^{\circ} C a n d {98.6}^{\circ} F \to 37^{\circ} C$
Hence, $Δ Q = m s Δ T = 80 \times 1000 \times (39 - 37)$
$16 \times 10^{4} c a l = 160 k c a l$