QBManager

Heat Transfer

149628 Two hot bodies $B_{1}$ and $B_{2}$ have temperatures $100^{\circ} C$ and $80^{\circ} C$ respectively time $t = 0$ . The temperature of the surrounding is $40^{\circ} C$ . The ratio of the respective rates of cooling $R_{1}$ and $R_{2}$ of the these two bodies at $t = 0$ is:

1

R_{1} : R_{2} = 3 : 2

2

R_{1} : R_{2} = 5 : 4

3

R_{1} : R_{2} = 2 : 3

4

R_{1} : R_{2} = 4 : 5

Explanation:

A Given,
Temperature of body $(B_{1}) = 100^{\circ} C$
Temperature of body $(B_{2}) = 80^{\circ} C$
Temperature of the surrounding $= 40^{\circ} C$
Rate of cooling for body $(B_{1}) = R_{1}$
Rate of cooling for body $(B_{2}) = R_{2}$ We know that,
Rate of cooling $(\frac{dQ}{dt}) = mc \frac{d θ}{dt}$
$\frac{dQ}{dt} \propto \frac{d θ}{dt}$
For body $B_{1}$ ,
$R_{1} = {(\frac{dQ}{dt})}_{1} \propto {(\frac{d θ}{dt})}_{1}$
For body $B_{2}$ ,
$R_{2} = {(\frac{dQ}{dt})}_{2} \propto {(\frac{d θ}{dt})}_{2}$
$\frac{R_{1}}{R_{2}} = \frac{(d θ)_{1}}{(d θ)_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{100 - 40}{80 - 40}$
$\frac{R_{1}}{R_{2}} = \frac{60}{40}$
$\frac{R_{1}}{R_{2}} = \frac{3}{2}$
$R_{1} : R_{2} = 3 : 2$

AP EAMCET-25.09.2020

Heat Transfer

149629 In Newton's cooling experiment, the water equivalent of two similar calorimeters is $10 g$ each. They are filled with $350 g$ of water and $300 g$ of a liquid (equal volumes) separately. The times taken by water and the liquid to cool from $70^{\circ} C$ to $60^{\circ} C$ are $3 \min$ and $95 \sec$ respectively. Then find the specific heat of the liquid.

1

0.3 cal . g^{- 1},^{\circ} C^{- 1}

2

0.5 cal . g^{- 1},^{\circ} C^{- 1}

3 0.6 cal.

g^{- 1},^{\circ} C^{- 1}

4

0.8 cal \cdot g^{- 1},^{\circ} C^{- 1}

Explanation:

C Given,
The water equivalent of two similar calorimeters
$(W) = 10 g each$
$1^{st}$ calorimeter filled with water of masses $(M_{w}) = 350 g$
$2^{nd}$ calorimeter filled with liquid of mass $(M_{ℓ}) = 300 g$
We know that,
Rate of heat loss of water $=$ Rate of heat loss of liquid
$\frac{(M_{w} C_{w} + W) (θ_{2} - θ_{1})}{t_{w}} = \frac{(M_{ℓ} C_{ℓ} + W) (θ_{2} - θ_{1})}{t_{ℓ}}$
$\frac{M_{w} C_{w} + W}{t_{w}} = \frac{M_{ℓ} C_{ℓ} + W}{t_{ℓ}}$
$C_{ℓ} = \frac{1}{m_{ℓ}} [(\frac{t_{ℓ}}{t_{w}}) (M_{w} C_{w} + W) - W]$
$C_{ℓ} = \frac{1}{300} [\frac{95}{180} {(350 \times 1) + 10} - 10]$
$= \frac{1}{300} [\frac{350 \times 95}{180} + \frac{95 \times 10}{180} - 10]$
$C_{ℓ} = 0.6 cal / {gm}^{\circ} C$

AP EAMCET-25.09.2020

Heat Transfer

149630 A body takes 10 minutes to cool from $60^{\circ} C$ to $50^{\circ} C$ . The temperature of same body after next 10 minutes will be of surroundings is $25^{\circ} C$ ) (If the temperature

1

40^{\circ} C

2

48^{\circ} C

3

43^{\circ} C

4

{45.5}^{\circ} X

Explanation:

C According of Newtons law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - 25)$
$\frac{60 - 50}{10} = K (\frac{60 + 50}{2} - 25)$
$1 = K . (55 - 25)$
$K = \frac{1}{30}$
Again according to Newtons law of cooling
$\frac{50 - θ}{t} = K (\frac{50 + θ}{2} - 25)$
$\frac{50 - θ}{10} = \frac{1}{30} (\frac{50 + θ}{2} - 25)$
$150 - 3 θ = (\frac{50 + θ - 50}{2})$
$300 - 6 θ = θ$
$7 θ = 300$
$θ = \frac{300}{7}$
$θ = {42.85}^{\circ} C$
$θ ≃ 43^{\circ} C$

AP EAMCET-07.10.2020

Heat Transfer

149631 When a meter rod made of silver at $0^{\circ} C$ is heated to $100^{\circ} C$ , its length increased by 0.19 cm. Then find the coefficient of volume expansion of the silver.

1

0.63 \times 10^{- 5}^{\circ} C^{- 1}

2

1.9 \times 10^{- 5}^{\circ} C^{- 1}

3

5.7 \times 10^{- 5} C^{- 1}

4

16.1 \times 10^{- 5}^{\circ} C^{- 1}

Explanation:

C Given,
Initial Temperature of $Rod = 0^{\circ}$
Final Temperature of $Rod = 100^{\circ} C$
Length increased $(Δ L) = 0.19 cm = 0.0019 m$
We know that,
Coefficient of linear thermal expansion $(α) = \frac{Δ l}{l Δ t}$
$α = \frac{0.0019}{1 \times 100}$
$α = 1.9 \times 10^{- 5}$
Coefficient of volume expansion $(γ) = 3 \times$ coefficient of linear expansion
$γ = 3 α$
$γ = 3 \times 1.9 \times 10^{- 5}$
$γ = 5.7 \times 10^{- 5}^{\circ} C^{- 1}$

AP EAMCET-24.08.2021

Heat Transfer

149632 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes. When the room temperature is at $20^{\circ} C$ , the time taken to $cool$ from $71^{\circ} C$ to $69^{\circ} C$ is.

1

50 \sec

2

40 \sec

3

38 \sec

4

42 \sec

Explanation:

D Given,
$θ_{1} = 94^{\circ} c θ_{2} = 86^{\circ} t = 2 \min θ_{0} = 20^{\circ} c$
For case 1
Newton's Law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
$\frac{94 - 86}{2} = K (\frac{94 + 86}{2} - 20)$
$4 = K \times 70$
For case 2
$θ_{1} = 71^{\circ} c θ_{2} = 69^{\circ} c θ_{0} = 20^{\circ}$
According to again Newton's Law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
$\frac{71 - 69}{t} = \frac{2}{35} (\frac{71 + 69}{2} - 20)$
$\frac{2}{t} = \frac{2}{35} \times 50$
$\frac{1}{t} = \frac{50}{35}$
$t = \frac{35}{50}$
$t = 0.7 min$
$t = 0.7 \times 60 \sec$
$t = 42 \sec$
The time taken to cool from $71^{\circ} C$ to $69^{\circ} C$ is $42 \sec$

AP EAMCET(Medical)-2015

Heat Transfer

149628 Two hot bodies $B_{1}$ and $B_{2}$ have temperatures $100^{\circ} C$ and $80^{\circ} C$ respectively time $t = 0$ . The temperature of the surrounding is $40^{\circ} C$ . The ratio of the respective rates of cooling $R_{1}$ and $R_{2}$ of the these two bodies at $t = 0$ is:

1

R_{1} : R_{2} = 3 : 2

2

R_{1} : R_{2} = 5 : 4

3

R_{1} : R_{2} = 2 : 3

4

R_{1} : R_{2} = 4 : 5

Explanation:

A Given,
Temperature of body $(B_{1}) = 100^{\circ} C$
Temperature of body $(B_{2}) = 80^{\circ} C$
Temperature of the surrounding $= 40^{\circ} C$
Rate of cooling for body $(B_{1}) = R_{1}$
Rate of cooling for body $(B_{2}) = R_{2}$ We know that,
Rate of cooling $(\frac{dQ}{dt}) = mc \frac{d θ}{dt}$
$\frac{dQ}{dt} \propto \frac{d θ}{dt}$
For body $B_{1}$ ,
$R_{1} = {(\frac{dQ}{dt})}_{1} \propto {(\frac{d θ}{dt})}_{1}$
For body $B_{2}$ ,
$R_{2} = {(\frac{dQ}{dt})}_{2} \propto {(\frac{d θ}{dt})}_{2}$
$\frac{R_{1}}{R_{2}} = \frac{(d θ)_{1}}{(d θ)_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{100 - 40}{80 - 40}$
$\frac{R_{1}}{R_{2}} = \frac{60}{40}$
$\frac{R_{1}}{R_{2}} = \frac{3}{2}$
$R_{1} : R_{2} = 3 : 2$

AP EAMCET-25.09.2020

Heat Transfer

149629 In Newton's cooling experiment, the water equivalent of two similar calorimeters is $10 g$ each. They are filled with $350 g$ of water and $300 g$ of a liquid (equal volumes) separately. The times taken by water and the liquid to cool from $70^{\circ} C$ to $60^{\circ} C$ are $3 \min$ and $95 \sec$ respectively. Then find the specific heat of the liquid.

1

0.3 cal . g^{- 1},^{\circ} C^{- 1}

2

0.5 cal . g^{- 1},^{\circ} C^{- 1}

3 0.6 cal.

g^{- 1},^{\circ} C^{- 1}

4

0.8 cal \cdot g^{- 1},^{\circ} C^{- 1}

Explanation:

C Given,
The water equivalent of two similar calorimeters
$(W) = 10 g each$
$1^{st}$ calorimeter filled with water of masses $(M_{w}) = 350 g$
$2^{nd}$ calorimeter filled with liquid of mass $(M_{ℓ}) = 300 g$
We know that,
Rate of heat loss of water $=$ Rate of heat loss of liquid
$\frac{(M_{w} C_{w} + W) (θ_{2} - θ_{1})}{t_{w}} = \frac{(M_{ℓ} C_{ℓ} + W) (θ_{2} - θ_{1})}{t_{ℓ}}$
$\frac{M_{w} C_{w} + W}{t_{w}} = \frac{M_{ℓ} C_{ℓ} + W}{t_{ℓ}}$
$C_{ℓ} = \frac{1}{m_{ℓ}} [(\frac{t_{ℓ}}{t_{w}}) (M_{w} C_{w} + W) - W]$
$C_{ℓ} = \frac{1}{300} [\frac{95}{180} {(350 \times 1) + 10} - 10]$
$= \frac{1}{300} [\frac{350 \times 95}{180} + \frac{95 \times 10}{180} - 10]$
$C_{ℓ} = 0.6 cal / {gm}^{\circ} C$

AP EAMCET-25.09.2020

Heat Transfer

149630 A body takes 10 minutes to cool from $60^{\circ} C$ to $50^{\circ} C$ . The temperature of same body after next 10 minutes will be of surroundings is $25^{\circ} C$ ) (If the temperature

1

40^{\circ} C

2

48^{\circ} C

3

43^{\circ} C

4

{45.5}^{\circ} X

Explanation:

C According of Newtons law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - 25)$
$\frac{60 - 50}{10} = K (\frac{60 + 50}{2} - 25)$
$1 = K . (55 - 25)$
$K = \frac{1}{30}$
Again according to Newtons law of cooling
$\frac{50 - θ}{t} = K (\frac{50 + θ}{2} - 25)$
$\frac{50 - θ}{10} = \frac{1}{30} (\frac{50 + θ}{2} - 25)$
$150 - 3 θ = (\frac{50 + θ - 50}{2})$
$300 - 6 θ = θ$
$7 θ = 300$
$θ = \frac{300}{7}$
$θ = {42.85}^{\circ} C$
$θ ≃ 43^{\circ} C$

AP EAMCET-07.10.2020

Heat Transfer

149631 When a meter rod made of silver at $0^{\circ} C$ is heated to $100^{\circ} C$ , its length increased by 0.19 cm. Then find the coefficient of volume expansion of the silver.

1

0.63 \times 10^{- 5}^{\circ} C^{- 1}

2

1.9 \times 10^{- 5}^{\circ} C^{- 1}

3

5.7 \times 10^{- 5} C^{- 1}

4

16.1 \times 10^{- 5}^{\circ} C^{- 1}

Explanation:

C Given,
Initial Temperature of $Rod = 0^{\circ}$
Final Temperature of $Rod = 100^{\circ} C$
Length increased $(Δ L) = 0.19 cm = 0.0019 m$
We know that,
Coefficient of linear thermal expansion $(α) = \frac{Δ l}{l Δ t}$
$α = \frac{0.0019}{1 \times 100}$
$α = 1.9 \times 10^{- 5}$
Coefficient of volume expansion $(γ) = 3 \times$ coefficient of linear expansion
$γ = 3 α$
$γ = 3 \times 1.9 \times 10^{- 5}$
$γ = 5.7 \times 10^{- 5}^{\circ} C^{- 1}$

AP EAMCET-24.08.2021

Heat Transfer

149632 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes. When the room temperature is at $20^{\circ} C$ , the time taken to $cool$ from $71^{\circ} C$ to $69^{\circ} C$ is.

1

50 \sec

2

40 \sec

3

38 \sec

4

42 \sec

Explanation:

D Given,
$θ_{1} = 94^{\circ} c θ_{2} = 86^{\circ} t = 2 \min θ_{0} = 20^{\circ} c$
For case 1
Newton's Law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
$\frac{94 - 86}{2} = K (\frac{94 + 86}{2} - 20)$
$4 = K \times 70$
For case 2
$θ_{1} = 71^{\circ} c θ_{2} = 69^{\circ} c θ_{0} = 20^{\circ}$
According to again Newton's Law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
$\frac{71 - 69}{t} = \frac{2}{35} (\frac{71 + 69}{2} - 20)$
$\frac{2}{t} = \frac{2}{35} \times 50$
$\frac{1}{t} = \frac{50}{35}$
$t = \frac{35}{50}$
$t = 0.7 min$
$t = 0.7 \times 60 \sec$
$t = 42 \sec$
The time taken to cool from $71^{\circ} C$ to $69^{\circ} C$ is $42 \sec$

AP EAMCET(Medical)-2015

Heat Transfer

149628 Two hot bodies $B_{1}$ and $B_{2}$ have temperatures $100^{\circ} C$ and $80^{\circ} C$ respectively time $t = 0$ . The temperature of the surrounding is $40^{\circ} C$ . The ratio of the respective rates of cooling $R_{1}$ and $R_{2}$ of the these two bodies at $t = 0$ is:

1

R_{1} : R_{2} = 3 : 2

2

R_{1} : R_{2} = 5 : 4

3

R_{1} : R_{2} = 2 : 3

4

R_{1} : R_{2} = 4 : 5

Explanation:

A Given,
Temperature of body $(B_{1}) = 100^{\circ} C$
Temperature of body $(B_{2}) = 80^{\circ} C$
Temperature of the surrounding $= 40^{\circ} C$
Rate of cooling for body $(B_{1}) = R_{1}$
Rate of cooling for body $(B_{2}) = R_{2}$ We know that,
Rate of cooling $(\frac{dQ}{dt}) = mc \frac{d θ}{dt}$
$\frac{dQ}{dt} \propto \frac{d θ}{dt}$
For body $B_{1}$ ,
$R_{1} = {(\frac{dQ}{dt})}_{1} \propto {(\frac{d θ}{dt})}_{1}$
For body $B_{2}$ ,
$R_{2} = {(\frac{dQ}{dt})}_{2} \propto {(\frac{d θ}{dt})}_{2}$
$\frac{R_{1}}{R_{2}} = \frac{(d θ)_{1}}{(d θ)_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{100 - 40}{80 - 40}$
$\frac{R_{1}}{R_{2}} = \frac{60}{40}$
$\frac{R_{1}}{R_{2}} = \frac{3}{2}$
$R_{1} : R_{2} = 3 : 2$

AP EAMCET-25.09.2020

Heat Transfer

149629 In Newton's cooling experiment, the water equivalent of two similar calorimeters is $10 g$ each. They are filled with $350 g$ of water and $300 g$ of a liquid (equal volumes) separately. The times taken by water and the liquid to cool from $70^{\circ} C$ to $60^{\circ} C$ are $3 \min$ and $95 \sec$ respectively. Then find the specific heat of the liquid.

1

0.3 cal . g^{- 1},^{\circ} C^{- 1}

2

0.5 cal . g^{- 1},^{\circ} C^{- 1}

3 0.6 cal.

g^{- 1},^{\circ} C^{- 1}

4

0.8 cal \cdot g^{- 1},^{\circ} C^{- 1}

Explanation:

C Given,
The water equivalent of two similar calorimeters
$(W) = 10 g each$
$1^{st}$ calorimeter filled with water of masses $(M_{w}) = 350 g$
$2^{nd}$ calorimeter filled with liquid of mass $(M_{ℓ}) = 300 g$
We know that,
Rate of heat loss of water $=$ Rate of heat loss of liquid
$\frac{(M_{w} C_{w} + W) (θ_{2} - θ_{1})}{t_{w}} = \frac{(M_{ℓ} C_{ℓ} + W) (θ_{2} - θ_{1})}{t_{ℓ}}$
$\frac{M_{w} C_{w} + W}{t_{w}} = \frac{M_{ℓ} C_{ℓ} + W}{t_{ℓ}}$
$C_{ℓ} = \frac{1}{m_{ℓ}} [(\frac{t_{ℓ}}{t_{w}}) (M_{w} C_{w} + W) - W]$
$C_{ℓ} = \frac{1}{300} [\frac{95}{180} {(350 \times 1) + 10} - 10]$
$= \frac{1}{300} [\frac{350 \times 95}{180} + \frac{95 \times 10}{180} - 10]$
$C_{ℓ} = 0.6 cal / {gm}^{\circ} C$

AP EAMCET-25.09.2020

Heat Transfer

149630 A body takes 10 minutes to cool from $60^{\circ} C$ to $50^{\circ} C$ . The temperature of same body after next 10 minutes will be of surroundings is $25^{\circ} C$ ) (If the temperature

1

40^{\circ} C

2

48^{\circ} C

3

43^{\circ} C

4

{45.5}^{\circ} X

Explanation:

C According of Newtons law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - 25)$
$\frac{60 - 50}{10} = K (\frac{60 + 50}{2} - 25)$
$1 = K . (55 - 25)$
$K = \frac{1}{30}$
Again according to Newtons law of cooling
$\frac{50 - θ}{t} = K (\frac{50 + θ}{2} - 25)$
$\frac{50 - θ}{10} = \frac{1}{30} (\frac{50 + θ}{2} - 25)$
$150 - 3 θ = (\frac{50 + θ - 50}{2})$
$300 - 6 θ = θ$
$7 θ = 300$
$θ = \frac{300}{7}$
$θ = {42.85}^{\circ} C$
$θ ≃ 43^{\circ} C$

AP EAMCET-07.10.2020

Heat Transfer

149631 When a meter rod made of silver at $0^{\circ} C$ is heated to $100^{\circ} C$ , its length increased by 0.19 cm. Then find the coefficient of volume expansion of the silver.

1

0.63 \times 10^{- 5}^{\circ} C^{- 1}

2

1.9 \times 10^{- 5}^{\circ} C^{- 1}

3

5.7 \times 10^{- 5} C^{- 1}

4

16.1 \times 10^{- 5}^{\circ} C^{- 1}

Explanation:

C Given,
Initial Temperature of $Rod = 0^{\circ}$
Final Temperature of $Rod = 100^{\circ} C$
Length increased $(Δ L) = 0.19 cm = 0.0019 m$
We know that,
Coefficient of linear thermal expansion $(α) = \frac{Δ l}{l Δ t}$
$α = \frac{0.0019}{1 \times 100}$
$α = 1.9 \times 10^{- 5}$
Coefficient of volume expansion $(γ) = 3 \times$ coefficient of linear expansion
$γ = 3 α$
$γ = 3 \times 1.9 \times 10^{- 5}$
$γ = 5.7 \times 10^{- 5}^{\circ} C^{- 1}$

AP EAMCET-24.08.2021

Heat Transfer

149632 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes. When the room temperature is at $20^{\circ} C$ , the time taken to $cool$ from $71^{\circ} C$ to $69^{\circ} C$ is.

1

50 \sec

2

40 \sec

3

38 \sec

4

42 \sec

Explanation:

D Given,
$θ_{1} = 94^{\circ} c θ_{2} = 86^{\circ} t = 2 \min θ_{0} = 20^{\circ} c$
For case 1
Newton's Law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
$\frac{94 - 86}{2} = K (\frac{94 + 86}{2} - 20)$
$4 = K \times 70$
For case 2
$θ_{1} = 71^{\circ} c θ_{2} = 69^{\circ} c θ_{0} = 20^{\circ}$
According to again Newton's Law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
$\frac{71 - 69}{t} = \frac{2}{35} (\frac{71 + 69}{2} - 20)$
$\frac{2}{t} = \frac{2}{35} \times 50$
$\frac{1}{t} = \frac{50}{35}$
$t = \frac{35}{50}$
$t = 0.7 min$
$t = 0.7 \times 60 \sec$
$t = 42 \sec$
The time taken to cool from $71^{\circ} C$ to $69^{\circ} C$ is $42 \sec$

AP EAMCET(Medical)-2015

Heat Transfer

149628 Two hot bodies $B_{1}$ and $B_{2}$ have temperatures $100^{\circ} C$ and $80^{\circ} C$ respectively time $t = 0$ . The temperature of the surrounding is $40^{\circ} C$ . The ratio of the respective rates of cooling $R_{1}$ and $R_{2}$ of the these two bodies at $t = 0$ is:

1

R_{1} : R_{2} = 3 : 2

2

R_{1} : R_{2} = 5 : 4

3

R_{1} : R_{2} = 2 : 3

4

R_{1} : R_{2} = 4 : 5

Explanation:

A Given,
Temperature of body $(B_{1}) = 100^{\circ} C$
Temperature of body $(B_{2}) = 80^{\circ} C$
Temperature of the surrounding $= 40^{\circ} C$
Rate of cooling for body $(B_{1}) = R_{1}$
Rate of cooling for body $(B_{2}) = R_{2}$ We know that,
Rate of cooling $(\frac{dQ}{dt}) = mc \frac{d θ}{dt}$
$\frac{dQ}{dt} \propto \frac{d θ}{dt}$
For body $B_{1}$ ,
$R_{1} = {(\frac{dQ}{dt})}_{1} \propto {(\frac{d θ}{dt})}_{1}$
For body $B_{2}$ ,
$R_{2} = {(\frac{dQ}{dt})}_{2} \propto {(\frac{d θ}{dt})}_{2}$
$\frac{R_{1}}{R_{2}} = \frac{(d θ)_{1}}{(d θ)_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{100 - 40}{80 - 40}$
$\frac{R_{1}}{R_{2}} = \frac{60}{40}$
$\frac{R_{1}}{R_{2}} = \frac{3}{2}$
$R_{1} : R_{2} = 3 : 2$

AP EAMCET-25.09.2020

Heat Transfer

149629 In Newton's cooling experiment, the water equivalent of two similar calorimeters is $10 g$ each. They are filled with $350 g$ of water and $300 g$ of a liquid (equal volumes) separately. The times taken by water and the liquid to cool from $70^{\circ} C$ to $60^{\circ} C$ are $3 \min$ and $95 \sec$ respectively. Then find the specific heat of the liquid.

1

0.3 cal . g^{- 1},^{\circ} C^{- 1}

2

0.5 cal . g^{- 1},^{\circ} C^{- 1}

3 0.6 cal.

g^{- 1},^{\circ} C^{- 1}

4

0.8 cal \cdot g^{- 1},^{\circ} C^{- 1}

Explanation:

C Given,
The water equivalent of two similar calorimeters
$(W) = 10 g each$
$1^{st}$ calorimeter filled with water of masses $(M_{w}) = 350 g$
$2^{nd}$ calorimeter filled with liquid of mass $(M_{ℓ}) = 300 g$
We know that,
Rate of heat loss of water $=$ Rate of heat loss of liquid
$\frac{(M_{w} C_{w} + W) (θ_{2} - θ_{1})}{t_{w}} = \frac{(M_{ℓ} C_{ℓ} + W) (θ_{2} - θ_{1})}{t_{ℓ}}$
$\frac{M_{w} C_{w} + W}{t_{w}} = \frac{M_{ℓ} C_{ℓ} + W}{t_{ℓ}}$
$C_{ℓ} = \frac{1}{m_{ℓ}} [(\frac{t_{ℓ}}{t_{w}}) (M_{w} C_{w} + W) - W]$
$C_{ℓ} = \frac{1}{300} [\frac{95}{180} {(350 \times 1) + 10} - 10]$
$= \frac{1}{300} [\frac{350 \times 95}{180} + \frac{95 \times 10}{180} - 10]$
$C_{ℓ} = 0.6 cal / {gm}^{\circ} C$

AP EAMCET-25.09.2020

Heat Transfer

149630 A body takes 10 minutes to cool from $60^{\circ} C$ to $50^{\circ} C$ . The temperature of same body after next 10 minutes will be of surroundings is $25^{\circ} C$ ) (If the temperature

1

40^{\circ} C

2

48^{\circ} C

3

43^{\circ} C

4

{45.5}^{\circ} X

Explanation:

C According of Newtons law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - 25)$
$\frac{60 - 50}{10} = K (\frac{60 + 50}{2} - 25)$
$1 = K . (55 - 25)$
$K = \frac{1}{30}$
Again according to Newtons law of cooling
$\frac{50 - θ}{t} = K (\frac{50 + θ}{2} - 25)$
$\frac{50 - θ}{10} = \frac{1}{30} (\frac{50 + θ}{2} - 25)$
$150 - 3 θ = (\frac{50 + θ - 50}{2})$
$300 - 6 θ = θ$
$7 θ = 300$
$θ = \frac{300}{7}$
$θ = {42.85}^{\circ} C$
$θ ≃ 43^{\circ} C$

AP EAMCET-07.10.2020

Heat Transfer

149631 When a meter rod made of silver at $0^{\circ} C$ is heated to $100^{\circ} C$ , its length increased by 0.19 cm. Then find the coefficient of volume expansion of the silver.

1

0.63 \times 10^{- 5}^{\circ} C^{- 1}

2

1.9 \times 10^{- 5}^{\circ} C^{- 1}

3

5.7 \times 10^{- 5} C^{- 1}

4

16.1 \times 10^{- 5}^{\circ} C^{- 1}

Explanation:

C Given,
Initial Temperature of $Rod = 0^{\circ}$
Final Temperature of $Rod = 100^{\circ} C$
Length increased $(Δ L) = 0.19 cm = 0.0019 m$
We know that,
Coefficient of linear thermal expansion $(α) = \frac{Δ l}{l Δ t}$
$α = \frac{0.0019}{1 \times 100}$
$α = 1.9 \times 10^{- 5}$
Coefficient of volume expansion $(γ) = 3 \times$ coefficient of linear expansion
$γ = 3 α$
$γ = 3 \times 1.9 \times 10^{- 5}$
$γ = 5.7 \times 10^{- 5}^{\circ} C^{- 1}$

AP EAMCET-24.08.2021

Heat Transfer

149632 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes. When the room temperature is at $20^{\circ} C$ , the time taken to $cool$ from $71^{\circ} C$ to $69^{\circ} C$ is.

1

50 \sec

2

40 \sec

3

38 \sec

4

42 \sec

Explanation:

D Given,
$θ_{1} = 94^{\circ} c θ_{2} = 86^{\circ} t = 2 \min θ_{0} = 20^{\circ} c$
For case 1
Newton's Law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
$\frac{94 - 86}{2} = K (\frac{94 + 86}{2} - 20)$
$4 = K \times 70$
For case 2
$θ_{1} = 71^{\circ} c θ_{2} = 69^{\circ} c θ_{0} = 20^{\circ}$
According to again Newton's Law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
$\frac{71 - 69}{t} = \frac{2}{35} (\frac{71 + 69}{2} - 20)$
$\frac{2}{t} = \frac{2}{35} \times 50$
$\frac{1}{t} = \frac{50}{35}$
$t = \frac{35}{50}$
$t = 0.7 min$
$t = 0.7 \times 60 \sec$
$t = 42 \sec$
The time taken to cool from $71^{\circ} C$ to $69^{\circ} C$ is $42 \sec$

AP EAMCET(Medical)-2015

Heat Transfer

149628 Two hot bodies $B_{1}$ and $B_{2}$ have temperatures $100^{\circ} C$ and $80^{\circ} C$ respectively time $t = 0$ . The temperature of the surrounding is $40^{\circ} C$ . The ratio of the respective rates of cooling $R_{1}$ and $R_{2}$ of the these two bodies at $t = 0$ is:

1

R_{1} : R_{2} = 3 : 2

2

R_{1} : R_{2} = 5 : 4

3

R_{1} : R_{2} = 2 : 3

4

R_{1} : R_{2} = 4 : 5

Explanation:

A Given,
Temperature of body $(B_{1}) = 100^{\circ} C$
Temperature of body $(B_{2}) = 80^{\circ} C$
Temperature of the surrounding $= 40^{\circ} C$
Rate of cooling for body $(B_{1}) = R_{1}$
Rate of cooling for body $(B_{2}) = R_{2}$ We know that,
Rate of cooling $(\frac{dQ}{dt}) = mc \frac{d θ}{dt}$
$\frac{dQ}{dt} \propto \frac{d θ}{dt}$
For body $B_{1}$ ,
$R_{1} = {(\frac{dQ}{dt})}_{1} \propto {(\frac{d θ}{dt})}_{1}$
For body $B_{2}$ ,
$R_{2} = {(\frac{dQ}{dt})}_{2} \propto {(\frac{d θ}{dt})}_{2}$
$\frac{R_{1}}{R_{2}} = \frac{(d θ)_{1}}{(d θ)_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{100 - 40}{80 - 40}$
$\frac{R_{1}}{R_{2}} = \frac{60}{40}$
$\frac{R_{1}}{R_{2}} = \frac{3}{2}$
$R_{1} : R_{2} = 3 : 2$

AP EAMCET-25.09.2020

Heat Transfer

149629 In Newton's cooling experiment, the water equivalent of two similar calorimeters is $10 g$ each. They are filled with $350 g$ of water and $300 g$ of a liquid (equal volumes) separately. The times taken by water and the liquid to cool from $70^{\circ} C$ to $60^{\circ} C$ are $3 \min$ and $95 \sec$ respectively. Then find the specific heat of the liquid.

1

0.3 cal . g^{- 1},^{\circ} C^{- 1}

2

0.5 cal . g^{- 1},^{\circ} C^{- 1}

3 0.6 cal.

g^{- 1},^{\circ} C^{- 1}

4

0.8 cal \cdot g^{- 1},^{\circ} C^{- 1}

Explanation:

C Given,
The water equivalent of two similar calorimeters
$(W) = 10 g each$
$1^{st}$ calorimeter filled with water of masses $(M_{w}) = 350 g$
$2^{nd}$ calorimeter filled with liquid of mass $(M_{ℓ}) = 300 g$
We know that,
Rate of heat loss of water $=$ Rate of heat loss of liquid
$\frac{(M_{w} C_{w} + W) (θ_{2} - θ_{1})}{t_{w}} = \frac{(M_{ℓ} C_{ℓ} + W) (θ_{2} - θ_{1})}{t_{ℓ}}$
$\frac{M_{w} C_{w} + W}{t_{w}} = \frac{M_{ℓ} C_{ℓ} + W}{t_{ℓ}}$
$C_{ℓ} = \frac{1}{m_{ℓ}} [(\frac{t_{ℓ}}{t_{w}}) (M_{w} C_{w} + W) - W]$
$C_{ℓ} = \frac{1}{300} [\frac{95}{180} {(350 \times 1) + 10} - 10]$
$= \frac{1}{300} [\frac{350 \times 95}{180} + \frac{95 \times 10}{180} - 10]$
$C_{ℓ} = 0.6 cal / {gm}^{\circ} C$

AP EAMCET-25.09.2020

Heat Transfer

149630 A body takes 10 minutes to cool from $60^{\circ} C$ to $50^{\circ} C$ . The temperature of same body after next 10 minutes will be of surroundings is $25^{\circ} C$ ) (If the temperature

1

40^{\circ} C

2

48^{\circ} C

3

43^{\circ} C

4

{45.5}^{\circ} X

Explanation:

C According of Newtons law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - 25)$
$\frac{60 - 50}{10} = K (\frac{60 + 50}{2} - 25)$
$1 = K . (55 - 25)$
$K = \frac{1}{30}$
Again according to Newtons law of cooling
$\frac{50 - θ}{t} = K (\frac{50 + θ}{2} - 25)$
$\frac{50 - θ}{10} = \frac{1}{30} (\frac{50 + θ}{2} - 25)$
$150 - 3 θ = (\frac{50 + θ - 50}{2})$
$300 - 6 θ = θ$
$7 θ = 300$
$θ = \frac{300}{7}$
$θ = {42.85}^{\circ} C$
$θ ≃ 43^{\circ} C$

AP EAMCET-07.10.2020

Heat Transfer

149631 When a meter rod made of silver at $0^{\circ} C$ is heated to $100^{\circ} C$ , its length increased by 0.19 cm. Then find the coefficient of volume expansion of the silver.

1

0.63 \times 10^{- 5}^{\circ} C^{- 1}

2

1.9 \times 10^{- 5}^{\circ} C^{- 1}

3

5.7 \times 10^{- 5} C^{- 1}

4

16.1 \times 10^{- 5}^{\circ} C^{- 1}

Explanation:

C Given,
Initial Temperature of $Rod = 0^{\circ}$
Final Temperature of $Rod = 100^{\circ} C$
Length increased $(Δ L) = 0.19 cm = 0.0019 m$
We know that,
Coefficient of linear thermal expansion $(α) = \frac{Δ l}{l Δ t}$
$α = \frac{0.0019}{1 \times 100}$
$α = 1.9 \times 10^{- 5}$
Coefficient of volume expansion $(γ) = 3 \times$ coefficient of linear expansion
$γ = 3 α$
$γ = 3 \times 1.9 \times 10^{- 5}$
$γ = 5.7 \times 10^{- 5}^{\circ} C^{- 1}$

AP EAMCET-24.08.2021

Heat Transfer

149632 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes. When the room temperature is at $20^{\circ} C$ , the time taken to $cool$ from $71^{\circ} C$ to $69^{\circ} C$ is.

1

50 \sec

2

40 \sec

3

38 \sec

4

42 \sec

Explanation:

D Given,
$θ_{1} = 94^{\circ} c θ_{2} = 86^{\circ} t = 2 \min θ_{0} = 20^{\circ} c$
For case 1
Newton's Law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
$\frac{94 - 86}{2} = K (\frac{94 + 86}{2} - 20)$
$4 = K \times 70$
For case 2
$θ_{1} = 71^{\circ} c θ_{2} = 69^{\circ} c θ_{0} = 20^{\circ}$
According to again Newton's Law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
$\frac{71 - 69}{t} = \frac{2}{35} (\frac{71 + 69}{2} - 20)$
$\frac{2}{t} = \frac{2}{35} \times 50$
$\frac{1}{t} = \frac{50}{35}$
$t = \frac{35}{50}$
$t = 0.7 min$
$t = 0.7 \times 60 \sec$
$t = 42 \sec$
The time taken to cool from $71^{\circ} C$ to $69^{\circ} C$ is $42 \sec$

AP EAMCET(Medical)-2015