Question

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 = 4.18 J/g°C).

Answer 1

Given m = 25 g, ∆T = (100 – 0) = 100°C Or in terms of Kelvin (373.15 – 273.15) = 100K, 

C = 4.18J/g°C 

Heat energy required, Q = m × C × ∆T = 25 × 4.18 × 100 = 10450 J