If the specific heat of lead is 0.03 cal/gm, then the thermal capacity of 500 gm of lead will be
A. `5cal//.^(@)C`
B. `10cal//.^(@)C`
C. `15cal//.^(@)C`
D. `20cal//.^(@)C`

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Correct Answer - C
`C=mc=500xx0.03=15cal//.^(@)C`

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