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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 Specific heat capacity of lead.
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Solution
Specific heat capacity of lead = `"Heat supplied"/("Mass" xx "Rise in temperature")`
`= (1560 "J")/((400 "g") xx (50^circ"C" - 20^circ "C")) = 0.13 "J"//"g"^circ "C"`
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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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