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The heat capacity of a vessel is 300 cal...

The heat capacity of a vessel is 300 cal `.^(@)C^(-1)` and the heat capacity of water contained in the vessel is also 300 cal `.^(@)C^(-1)`. How much heat (in joules) is required to raise the temperature of water in the vessel by `126^(@)F`?

Text Solution

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A rise in temeprature of `9^(@)C` is equal to a rise in temperature of `5.^(@)C`.
`therefore`A rise in temperature of 126 `.^(@)F` is on the celsius scale equivalent to
`126xx(5)/(9)=14xx5=70.^(@)C`
quantity of heat required to raise the temperature of the vessel by `1.^(@)C=300cal` ltBrgt `therefore` Heat requried to raise its temperature by ltBrgt `70.^(@)C=70xx300=21000` cal.
similarly heat required to raise the temperature of water by `70.^(@)C=70xx300=21000cal`
`therefore` the total heat required `=(21000+21000)cal=42000cal`
`=4200xx4.2=176.4kJ`
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