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 ).
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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1 Answer
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11.13 Mass of the copper block, m = 2.5 kg = 2.5 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 = 487500 J
Let gm be the mass of the ice, which will melt because of the copper block.
Heat gained by ice block = Q =
= g = 1455.22 gm = 1.45 kg
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According to question, we can write
Heat Released by block = Heat gain by large Ice block
5 × 0.39 × 500 = mice × 335
= 2.91 kg
Given
6a2 = 24
a2 = 4
a = 2m
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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(a), (d) During the process AB temperature of the system is 00C . hence it represents phase change that is transformation of ice into water while temperature remains 00C.
BC represents rise in temperature of water from 00C to 1000C. now water starts converting into steam which is represented by CD.
Which is only represents by a, d options
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