Question

If you wish to warm 100 kg of water by 20°C for your bath, how much heat is required? (Give your answer in calories and joules.)

Answer 1

Q = mcθ

Where m = mass of water in kg.
c = specific heat capacity in kJ/kg⁰C, c for water = 4200 kJ/kg⁰C
θ = temperature rise in ⁰C

Q = 100*4200* 20 Note here the temperature rise is 20 ⁰C
Q = 8 400 000 J

In calories, 4.2 J = 1 Calorie
= 8 400 000 / 4.2 = 200 000

Q = 200 000 Calories