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physics ncert solutions class 11th Physics Class 11th Thermal Properties of Matter

11.13 A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g^–1 K^–1; heat of fusion of water = 335 J g^–1 ).

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Payal Gupta | Contributor-Level 10

5 months ago

11.13 Mass of the copper block, m = 2.5 kg = 2.5 $\times 10^{3}$ gm

Rise in temperature of the copper block, $?$ T = 500°C

Specific heat of the copper, C = 0.39 g–1 K–1

Heat of fusion of water, L = 335 J g–1

The maximum heat the copper block can lose, Q = mc $?$ T = 2.5 $\times 10^{3} \times 0.39 \times 500$ = 487500 J

Let $m_{1}$ gm be the mass of the ice, which will melt because of the copper block.

Heat gained by ice block = Q = $m_{1} L$

$m_{1} = \frac{Q}{L}$ = $\frac{487500}{335}$ g = 1455.22 gm = 1.45 kg

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A copper block of mass 5.0kg is heated to a temperature of $500 ° C$ and is placed on a large ice block. What is the maximum amount of ice that can melt?

[Specific heat of copper : 0.39 J g^-1 ⁰C^-1 and latent heat of fusion of water : 335 J g^-1]

Raj Pandey

Heat Released by block = Heat gain by large Ice block

$\Rightarrow m c Δ T = M_{i c e} L$

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$m_{i c e} = \frac{5 \times 0.39 \times 500}{335}$

= 2.91 kg

A solid metallic cube having total surface area 24 m² is uniformly heated. if its temperature is increased by 10⁰C, calculate the increase in volume of the cube.

(Given a = 5.0 × 10^-4 ⁰C^-1).

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Given

6a² = 24

a2 = 4

a = 2m

$Δ v = v (γ) Δ T$

$= 3 α v Δ T$

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$= 3 \times 5 \times 1 0^{- 4} \times 8 \times 10$

$\begin{array}{l} = 120 \times 1 0^{3} \\ = 1.2 \times 1 0^{5} c m^{3} \end{array}$

A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct?

(a) The rate of cooling is constant till milk attains the temperature of the surrounding

(b) The temperature of milk falls off exponentially with time

(c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools

(d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk to the surroundings

Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c), (d) When the hot milk in the table is transferred to the surroundings by conduction, convection and radiation.

According to newton's law of cooling temperature of the milk falls of exponentially. Heat also will be transferred from surroundings to the milk but will be lesser than that of transferred from milk to surroundings. So option b, c, d satisfy.

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11.13 A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g^–1 K^–1; heat of fusion of water = 335 J g^–1 ).

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11.13 A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g–1 K–1; heat of fusion of water = 335 J g–1 ).

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11.13 A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g^–1 K^–1; heat of fusion of water = 335 J g^–1 ).