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Thermodynamics

148215 A $10 H$ inductor carries of $20 A$ . If the energy stored in this inductor melts ice at $0^{\circ} C$ , then the mass of the ice melted is [Latent heat of ice $= 2.26 \times 10^{3} J {kg}^{- 1}]$

1

0.6 kg

2

0.88 kg

3

0.98 kg

4

1.4 kg

Explanation:

B Given that,
Inductance $= 10$ Henry

Current $= 20 A$
Temperature $Δ t = 0^{\circ} C$
Latent heat of fusion $= 2.26 \times 10^{3} J / kg$
As we know that,
Energy stores $= \frac{1}{2} L_{i}^{2}$
This stored energy is fully used in melting the ice
$Q = {mL}_{f}$
Hence, $\frac{1}{2} L_{i}^{2} = {mL}_{f}$
Mass of melted ice $(m) = \frac{1}{2} \frac{L_{i}^{2}}{L_{f}}$
$= \frac{10 \times (20)^{2}}{2 . (2.26 \times 10^{3})}$
$m = \frac{10 \times 400}{4.52 \times 10^{3}}$
$= \frac{4}{4.52}$
$= 0.8849$
$= 0.88 kg$

AP EAMCET (Medical)-24.04.2019

Thermodynamics

148216 A small electric heater is used to heat $200 g$ of water. The time required to bring all this water from $40^{\circ} C$ to $100^{\circ} C$ is $200 s$ . If specific heat of the water is $42 {Jkg}^{- 1} K^{- 1}$ then the power supplied by the heater is-

1

155 W

2

310 W

3

88 W

4

252 W

Explanation:

D Given that,
Mass of water $= 200 gm$
Initial temperature of water $= 40^{\circ} C$
Final temperature $= 100^{\circ} C$
Specific heat of water $= 42 J / kg k$
Time $= 200$ Sec.
Heat energy, $Q = ms Δ t$
$= 20 \times 42 \times (100 - 40)$
$Q = 840 \times 60$
$Q = 50400$
$Power, P = \frac{Q}{t}$
$= \frac{50400}{200}$
$= 252 watt$

AP EAMCET-08.07.2022

Thermodynamics

148217 $5 g$ of ice at $- 30^{\circ} C$ and $20 g$ of water at $35^{\circ} C$ are mixed together in a calorimeter. The final temperature of the mixture is
(Neglect heat capacity of the calorimeter specific hear capacity of ice $= 0.5 cal g^{- 1}^{0} C^{- 1}$ and latent heat of fusion of ice $= 80 cal g^{- 1}$ and specific hear capacity of water $= 1 cal g^{- 1}^{\circ} C^{- 1}$ )

1

0^{\circ} C

2

4^{\circ} C

3

5^{\circ} C

4

9^{\circ} C

Explanation:

D Consider the final temperature of mixture be $T$ Given,
Water mass, $m = 20 g$ at $35^{\circ} C$
Ice mass $m = 5 g$ at $30^{\circ} C$
Heat lost by water, $Q_{1} = 20 \times 1 \times (35 - T)$
Heat taken by ice, $Q_{2} = {mC}_{i} Δ T + mL + {mC}_{w} T$
$= (5 \times 0.5 \times 30) + (5 \times 80) + 5 \times 1 \times T$
$= 75 + 400 + 5 T$
equation (i) $=$ equation (ii) $20 \times (35 - T) = 475 + 5 T$
$20 \times 35 - 20 T = 475 + 5 T$
$700 - 20 T = 475 + 5 T$
$225 = 25 T$
$T = 9^{\circ} C$

AP EAMCET-05.07.2022

Thermodynamics

148218 If a $4 kg$ of ice is inside a closed cubical thermacol box of side length $20 cm$ and wall thickness $4 cm$ then the mass of the ice remaining after 10 hours is nearly
(The out side temperature $= 50^{\circ} C$ coefficient of thermal conductivity of thermacol $= 0.01 {Js}^{- 1} m^{- 1 \circ} C^{- 1}$
Latent heat of fusion of ice $= 335 \times 10^{3} J {kg}^{- 1}$ )

1

3.678 kg

2

6.378 kg

3

2.87 kg

4

1.87 kg

Explanation:

A Given,
thermal conductivity,
$K = 0.01 {Js}^{- 1} m^{- 1}^{\circ} C^{- 1}$
Latent heat $= 335 \times 10^{3} J / kg$
Mass of ice $m_{i} = 4 kg$
Thickness of box, $b = 4 cm$
$= 4 \times 10^{- 2} m$
side length of box, $l = 20 \times 10^{- 2} m$
Total surface area of box $= 6 l^{2}$
$= 6 \times {(20 \times 10^{- 2})}^{2}$
$= 24 \times 10^{- 2} m^{2}$
Change in temperature
$Δ T = (50^{\circ} C - 0)$
$= 50^{\circ} C$
Total heat, $Q = \frac{KA Δ T}{b} \times$ time
$= \frac{0.01 \times 24 \times 10^{- 2} \times 50}{4 \times 10^{- 2}} \times (10 \times 60 \times 60)$
$= \frac{1200 \times 0.01}{4} \times 36 \times 10^{3}$
$= 3 \times 36 \times 10^{3}$
$= 108 \times 10^{3}$
Melt ice mass,
$Heat, Q = mL$
$mass (m) = \frac{Q}{L} = \frac{108 \times 10^{3}}{335 \times 10^{3}} = 0.322 kg$
Now, the mass of remaining part of ice $= (4 - 0.322)$
$= 3.6776$
$= 3.678 kg$

AP EAMCET-25.04.2017

Thermodynamics

148215 A $10 H$ inductor carries of $20 A$ . If the energy stored in this inductor melts ice at $0^{\circ} C$ , then the mass of the ice melted is [Latent heat of ice $= 2.26 \times 10^{3} J {kg}^{- 1}]$

1

0.6 kg

2

0.88 kg

3

0.98 kg

4

1.4 kg

Explanation:

B Given that,
Inductance $= 10$ Henry

Current $= 20 A$
Temperature $Δ t = 0^{\circ} C$
Latent heat of fusion $= 2.26 \times 10^{3} J / kg$
As we know that,
Energy stores $= \frac{1}{2} L_{i}^{2}$
This stored energy is fully used in melting the ice
$Q = {mL}_{f}$
Hence, $\frac{1}{2} L_{i}^{2} = {mL}_{f}$
Mass of melted ice $(m) = \frac{1}{2} \frac{L_{i}^{2}}{L_{f}}$
$= \frac{10 \times (20)^{2}}{2 . (2.26 \times 10^{3})}$
$m = \frac{10 \times 400}{4.52 \times 10^{3}}$
$= \frac{4}{4.52}$
$= 0.8849$
$= 0.88 kg$

AP EAMCET (Medical)-24.04.2019

Thermodynamics

148216 A small electric heater is used to heat $200 g$ of water. The time required to bring all this water from $40^{\circ} C$ to $100^{\circ} C$ is $200 s$ . If specific heat of the water is $42 {Jkg}^{- 1} K^{- 1}$ then the power supplied by the heater is-

1

155 W

2

310 W

3

88 W

4

252 W

Explanation:

D Given that,
Mass of water $= 200 gm$
Initial temperature of water $= 40^{\circ} C$
Final temperature $= 100^{\circ} C$
Specific heat of water $= 42 J / kg k$
Time $= 200$ Sec.
Heat energy, $Q = ms Δ t$
$= 20 \times 42 \times (100 - 40)$
$Q = 840 \times 60$
$Q = 50400$
$Power, P = \frac{Q}{t}$
$= \frac{50400}{200}$
$= 252 watt$

AP EAMCET-08.07.2022

Thermodynamics

148217 $5 g$ of ice at $- 30^{\circ} C$ and $20 g$ of water at $35^{\circ} C$ are mixed together in a calorimeter. The final temperature of the mixture is
(Neglect heat capacity of the calorimeter specific hear capacity of ice $= 0.5 cal g^{- 1}^{0} C^{- 1}$ and latent heat of fusion of ice $= 80 cal g^{- 1}$ and specific hear capacity of water $= 1 cal g^{- 1}^{\circ} C^{- 1}$ )

1

0^{\circ} C

2

4^{\circ} C

3

5^{\circ} C

4

9^{\circ} C

Explanation:

D Consider the final temperature of mixture be $T$ Given,
Water mass, $m = 20 g$ at $35^{\circ} C$
Ice mass $m = 5 g$ at $30^{\circ} C$
Heat lost by water, $Q_{1} = 20 \times 1 \times (35 - T)$
Heat taken by ice, $Q_{2} = {mC}_{i} Δ T + mL + {mC}_{w} T$
$= (5 \times 0.5 \times 30) + (5 \times 80) + 5 \times 1 \times T$
$= 75 + 400 + 5 T$
equation (i) $=$ equation (ii) $20 \times (35 - T) = 475 + 5 T$
$20 \times 35 - 20 T = 475 + 5 T$
$700 - 20 T = 475 + 5 T$
$225 = 25 T$
$T = 9^{\circ} C$

AP EAMCET-05.07.2022

Thermodynamics

148218 If a $4 kg$ of ice is inside a closed cubical thermacol box of side length $20 cm$ and wall thickness $4 cm$ then the mass of the ice remaining after 10 hours is nearly
(The out side temperature $= 50^{\circ} C$ coefficient of thermal conductivity of thermacol $= 0.01 {Js}^{- 1} m^{- 1 \circ} C^{- 1}$
Latent heat of fusion of ice $= 335 \times 10^{3} J {kg}^{- 1}$ )

1

3.678 kg

2

6.378 kg

3

2.87 kg

4

1.87 kg

Explanation:

A Given,
thermal conductivity,
$K = 0.01 {Js}^{- 1} m^{- 1}^{\circ} C^{- 1}$
Latent heat $= 335 \times 10^{3} J / kg$
Mass of ice $m_{i} = 4 kg$
Thickness of box, $b = 4 cm$
$= 4 \times 10^{- 2} m$
side length of box, $l = 20 \times 10^{- 2} m$
Total surface area of box $= 6 l^{2}$
$= 6 \times {(20 \times 10^{- 2})}^{2}$
$= 24 \times 10^{- 2} m^{2}$
Change in temperature
$Δ T = (50^{\circ} C - 0)$
$= 50^{\circ} C$
Total heat, $Q = \frac{KA Δ T}{b} \times$ time
$= \frac{0.01 \times 24 \times 10^{- 2} \times 50}{4 \times 10^{- 2}} \times (10 \times 60 \times 60)$
$= \frac{1200 \times 0.01}{4} \times 36 \times 10^{3}$
$= 3 \times 36 \times 10^{3}$
$= 108 \times 10^{3}$
Melt ice mass,
$Heat, Q = mL$
$mass (m) = \frac{Q}{L} = \frac{108 \times 10^{3}}{335 \times 10^{3}} = 0.322 kg$
Now, the mass of remaining part of ice $= (4 - 0.322)$
$= 3.6776$
$= 3.678 kg$

AP EAMCET-25.04.2017

Thermodynamics

148215 A $10 H$ inductor carries of $20 A$ . If the energy stored in this inductor melts ice at $0^{\circ} C$ , then the mass of the ice melted is [Latent heat of ice $= 2.26 \times 10^{3} J {kg}^{- 1}]$

1

0.6 kg

2

0.88 kg

3

0.98 kg

4

1.4 kg

Explanation:

B Given that,
Inductance $= 10$ Henry

Current $= 20 A$
Temperature $Δ t = 0^{\circ} C$
Latent heat of fusion $= 2.26 \times 10^{3} J / kg$
As we know that,
Energy stores $= \frac{1}{2} L_{i}^{2}$
This stored energy is fully used in melting the ice
$Q = {mL}_{f}$
Hence, $\frac{1}{2} L_{i}^{2} = {mL}_{f}$
Mass of melted ice $(m) = \frac{1}{2} \frac{L_{i}^{2}}{L_{f}}$
$= \frac{10 \times (20)^{2}}{2 . (2.26 \times 10^{3})}$
$m = \frac{10 \times 400}{4.52 \times 10^{3}}$
$= \frac{4}{4.52}$
$= 0.8849$
$= 0.88 kg$

AP EAMCET (Medical)-24.04.2019

Thermodynamics

148216 A small electric heater is used to heat $200 g$ of water. The time required to bring all this water from $40^{\circ} C$ to $100^{\circ} C$ is $200 s$ . If specific heat of the water is $42 {Jkg}^{- 1} K^{- 1}$ then the power supplied by the heater is-

1

155 W

2

310 W

3

88 W

4

252 W

Explanation:

D Given that,
Mass of water $= 200 gm$
Initial temperature of water $= 40^{\circ} C$
Final temperature $= 100^{\circ} C$
Specific heat of water $= 42 J / kg k$
Time $= 200$ Sec.
Heat energy, $Q = ms Δ t$
$= 20 \times 42 \times (100 - 40)$
$Q = 840 \times 60$
$Q = 50400$
$Power, P = \frac{Q}{t}$
$= \frac{50400}{200}$
$= 252 watt$

AP EAMCET-08.07.2022

Thermodynamics

148217 $5 g$ of ice at $- 30^{\circ} C$ and $20 g$ of water at $35^{\circ} C$ are mixed together in a calorimeter. The final temperature of the mixture is
(Neglect heat capacity of the calorimeter specific hear capacity of ice $= 0.5 cal g^{- 1}^{0} C^{- 1}$ and latent heat of fusion of ice $= 80 cal g^{- 1}$ and specific hear capacity of water $= 1 cal g^{- 1}^{\circ} C^{- 1}$ )

1

0^{\circ} C

2

4^{\circ} C

3

5^{\circ} C

4

9^{\circ} C

Explanation:

D Consider the final temperature of mixture be $T$ Given,
Water mass, $m = 20 g$ at $35^{\circ} C$
Ice mass $m = 5 g$ at $30^{\circ} C$
Heat lost by water, $Q_{1} = 20 \times 1 \times (35 - T)$
Heat taken by ice, $Q_{2} = {mC}_{i} Δ T + mL + {mC}_{w} T$
$= (5 \times 0.5 \times 30) + (5 \times 80) + 5 \times 1 \times T$
$= 75 + 400 + 5 T$
equation (i) $=$ equation (ii) $20 \times (35 - T) = 475 + 5 T$
$20 \times 35 - 20 T = 475 + 5 T$
$700 - 20 T = 475 + 5 T$
$225 = 25 T$
$T = 9^{\circ} C$

AP EAMCET-05.07.2022

Thermodynamics

148218 If a $4 kg$ of ice is inside a closed cubical thermacol box of side length $20 cm$ and wall thickness $4 cm$ then the mass of the ice remaining after 10 hours is nearly
(The out side temperature $= 50^{\circ} C$ coefficient of thermal conductivity of thermacol $= 0.01 {Js}^{- 1} m^{- 1 \circ} C^{- 1}$
Latent heat of fusion of ice $= 335 \times 10^{3} J {kg}^{- 1}$ )

1

3.678 kg

2

6.378 kg

3

2.87 kg

4

1.87 kg

Explanation:

A Given,
thermal conductivity,
$K = 0.01 {Js}^{- 1} m^{- 1}^{\circ} C^{- 1}$
Latent heat $= 335 \times 10^{3} J / kg$
Mass of ice $m_{i} = 4 kg$
Thickness of box, $b = 4 cm$
$= 4 \times 10^{- 2} m$
side length of box, $l = 20 \times 10^{- 2} m$
Total surface area of box $= 6 l^{2}$
$= 6 \times {(20 \times 10^{- 2})}^{2}$
$= 24 \times 10^{- 2} m^{2}$
Change in temperature
$Δ T = (50^{\circ} C - 0)$
$= 50^{\circ} C$
Total heat, $Q = \frac{KA Δ T}{b} \times$ time
$= \frac{0.01 \times 24 \times 10^{- 2} \times 50}{4 \times 10^{- 2}} \times (10 \times 60 \times 60)$
$= \frac{1200 \times 0.01}{4} \times 36 \times 10^{3}$
$= 3 \times 36 \times 10^{3}$
$= 108 \times 10^{3}$
Melt ice mass,
$Heat, Q = mL$
$mass (m) = \frac{Q}{L} = \frac{108 \times 10^{3}}{335 \times 10^{3}} = 0.322 kg$
Now, the mass of remaining part of ice $= (4 - 0.322)$
$= 3.6776$
$= 3.678 kg$

AP EAMCET-25.04.2017

Thermodynamics

148215 A $10 H$ inductor carries of $20 A$ . If the energy stored in this inductor melts ice at $0^{\circ} C$ , then the mass of the ice melted is [Latent heat of ice $= 2.26 \times 10^{3} J {kg}^{- 1}]$

1

0.6 kg

2

0.88 kg

3

0.98 kg

4

1.4 kg

Explanation:

B Given that,
Inductance $= 10$ Henry

Current $= 20 A$
Temperature $Δ t = 0^{\circ} C$
Latent heat of fusion $= 2.26 \times 10^{3} J / kg$
As we know that,
Energy stores $= \frac{1}{2} L_{i}^{2}$
This stored energy is fully used in melting the ice
$Q = {mL}_{f}$
Hence, $\frac{1}{2} L_{i}^{2} = {mL}_{f}$
Mass of melted ice $(m) = \frac{1}{2} \frac{L_{i}^{2}}{L_{f}}$
$= \frac{10 \times (20)^{2}}{2 . (2.26 \times 10^{3})}$
$m = \frac{10 \times 400}{4.52 \times 10^{3}}$
$= \frac{4}{4.52}$
$= 0.8849$
$= 0.88 kg$

AP EAMCET (Medical)-24.04.2019

Thermodynamics

148216 A small electric heater is used to heat $200 g$ of water. The time required to bring all this water from $40^{\circ} C$ to $100^{\circ} C$ is $200 s$ . If specific heat of the water is $42 {Jkg}^{- 1} K^{- 1}$ then the power supplied by the heater is-

1

155 W

2

310 W

3

88 W

4

252 W

Explanation:

D Given that,
Mass of water $= 200 gm$
Initial temperature of water $= 40^{\circ} C$
Final temperature $= 100^{\circ} C$
Specific heat of water $= 42 J / kg k$
Time $= 200$ Sec.
Heat energy, $Q = ms Δ t$
$= 20 \times 42 \times (100 - 40)$
$Q = 840 \times 60$
$Q = 50400$
$Power, P = \frac{Q}{t}$
$= \frac{50400}{200}$
$= 252 watt$

AP EAMCET-08.07.2022

Thermodynamics

148217 $5 g$ of ice at $- 30^{\circ} C$ and $20 g$ of water at $35^{\circ} C$ are mixed together in a calorimeter. The final temperature of the mixture is
(Neglect heat capacity of the calorimeter specific hear capacity of ice $= 0.5 cal g^{- 1}^{0} C^{- 1}$ and latent heat of fusion of ice $= 80 cal g^{- 1}$ and specific hear capacity of water $= 1 cal g^{- 1}^{\circ} C^{- 1}$ )

1

0^{\circ} C

2

4^{\circ} C

3

5^{\circ} C

4

9^{\circ} C

Explanation:

D Consider the final temperature of mixture be $T$ Given,
Water mass, $m = 20 g$ at $35^{\circ} C$
Ice mass $m = 5 g$ at $30^{\circ} C$
Heat lost by water, $Q_{1} = 20 \times 1 \times (35 - T)$
Heat taken by ice, $Q_{2} = {mC}_{i} Δ T + mL + {mC}_{w} T$
$= (5 \times 0.5 \times 30) + (5 \times 80) + 5 \times 1 \times T$
$= 75 + 400 + 5 T$
equation (i) $=$ equation (ii) $20 \times (35 - T) = 475 + 5 T$
$20 \times 35 - 20 T = 475 + 5 T$
$700 - 20 T = 475 + 5 T$
$225 = 25 T$
$T = 9^{\circ} C$

AP EAMCET-05.07.2022

Thermodynamics

148218 If a $4 kg$ of ice is inside a closed cubical thermacol box of side length $20 cm$ and wall thickness $4 cm$ then the mass of the ice remaining after 10 hours is nearly
(The out side temperature $= 50^{\circ} C$ coefficient of thermal conductivity of thermacol $= 0.01 {Js}^{- 1} m^{- 1 \circ} C^{- 1}$
Latent heat of fusion of ice $= 335 \times 10^{3} J {kg}^{- 1}$ )

1

3.678 kg

2

6.378 kg

3

2.87 kg

4

1.87 kg

Explanation:

A Given,
thermal conductivity,
$K = 0.01 {Js}^{- 1} m^{- 1}^{\circ} C^{- 1}$
Latent heat $= 335 \times 10^{3} J / kg$
Mass of ice $m_{i} = 4 kg$
Thickness of box, $b = 4 cm$
$= 4 \times 10^{- 2} m$
side length of box, $l = 20 \times 10^{- 2} m$
Total surface area of box $= 6 l^{2}$
$= 6 \times {(20 \times 10^{- 2})}^{2}$
$= 24 \times 10^{- 2} m^{2}$
Change in temperature
$Δ T = (50^{\circ} C - 0)$
$= 50^{\circ} C$
Total heat, $Q = \frac{KA Δ T}{b} \times$ time
$= \frac{0.01 \times 24 \times 10^{- 2} \times 50}{4 \times 10^{- 2}} \times (10 \times 60 \times 60)$
$= \frac{1200 \times 0.01}{4} \times 36 \times 10^{3}$
$= 3 \times 36 \times 10^{3}$
$= 108 \times 10^{3}$
Melt ice mass,
$Heat, Q = mL$
$mass (m) = \frac{Q}{L} = \frac{108 \times 10^{3}}{335 \times 10^{3}} = 0.322 kg$
Now, the mass of remaining part of ice $= (4 - 0.322)$
$= 3.6776$
$= 3.678 kg$

AP EAMCET-25.04.2017