Question

Explain how you would calculate the q for warming 100.0 grams of liquid water from 0°C to 100 °C.

Answer 1

Given:-

mass of water, m=100 g=0.1 kg

$\text{Initial temperature, }T_{i}=0^{0}C\\\text{Final temperature, }T_{f}=100^{0}C$

To Find:-

amount of heat transfer

Explanation:-

For liquid water, specific heat(C)=4.18 kJ/kg-K

Since, temperature of water increases from 0^{C} to 100^{0}C. Therefore, it is a case of specific heat.

$\Rightarrow q=mC\times(T_{f}-T_{i})\\\Rightarrow q=0.1\times4.18\times(100-0)=41.8kJ\\\text{Heat trequired to raise the temperature of water from } 0^{0}C \text{ to } 100^{0}C \text{ is }41.8kJ$