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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?

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प्रश्न

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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उत्तर १

Mass of the copper block, = 2.5 kg = 2500 g

Rise in the temperature of the copper block, Δθ = 500°C

Specific heat of copper, C = 0.39 J g–1 °C–1

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

The maximum heat the copper block can lose, Q = mCΔθ

= 2500 × 0.39 × 500

= 487500 J

Let m1 g be the amount of ice that melts when the copper block is placed on the ice block.

The heat gained by the melted ice, Q = m1L

`:.m_1 =Q/L = 487500/335 = 1455.22 g`

Hence, the maximum amount of ice that can melt is 1.45 kg.

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उत्तर २

Here, mass of copper block, m = 2.5 kg = 2500 g

Fall in temperature, `triangle T = 500 - 0= 500^@C`

Specific heat of copper, `c = 0.39 J g^(-1) ""^@C^(-1)`

Latent heat of fusion, `L = 335 J g^(-1)`

Let the mass of ice melted be m'

As, Heat gained by ice = Heat lost by copper

`:. m'L = mc triangle T`

`m' = (mc triangleT)/L`

`m' = (2500xx0.39xx500)/335 =1500 g = 1.5 kg`

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पाठ 10: Thermal Properties of Matter - Exercises [पृष्ठ २९५]

APPEARS IN

एनसीईआरटी Physics Part 1 and 2 [English] Class 11
पाठ 10 Thermal Properties of Matter
Exercises | Q 13 | पृष्ठ २९५

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