Numerical Problem.
What is the heat in joules required to raise the temperature of 25 grams of water from 0°C to 100°C? What is the heat in Calories? (Specific heat of water $=\frac{4.18 J }{ g ^{\circ} C }$
Or in terms of Kelvin $(373.15-273.15)=100 K$,
$C =\frac{4.18 J }{ g ^{\circ} C }$
Heat energy required, Q = m × C × ∆T = 25 × 4.18 × 100 = 10450 J