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Question

If the specific heat of lead is 0.03calg−1oC−1, thermal capacity of 500 g of lead is

A
5caloC1
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B
10caloC1
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C
15caloC1
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D
20caloC1
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Solution

The correct option is C 15caloC1
Given S=0.03calg1oC1
m=500g
C=mS=500×0.03=15caloC1

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