Question

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 $126^(@)F$?

Answer 1

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 $=(21000+21000)cal=42000cal$
$=4200xx4.2=176.4kJ$