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Question
The temperature of a lead piece of mass 400 g rises from 20°C to 50°C when 1560 J of heat is supplied to it. Calculate: Heat capacity of lead piece.
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Solution
Heat capacity of lead piece
= `"Heat supplied"/"Rise in temperature" = (1560 "J")/(50 - 20^circ "C")` = 52 J/° C.
RELATED QUESTIONS
Given below are observations on molar specific heats at room temperature of some common gases.
| Gas |
Molar specific heat (Cv) (cal mol–1 K–1) |
| Hydrogen | 4.87 |
| Nitrogen | 4.97 |
| Oxygen | 5.02 |
| Nitric oxide | 4.99 |
| Carbon monoxide | 5.01 |
| Chlorine | 6.17 |
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