**Given** Mass of water = 10 kg, Initial temperature = 90^{0} C = 363 K

Final temperature = 100^{0} C = 373 K, Specific heat of water = 4180 J/kg K

**Asked** Change in entropy, ΔS.

**Formulae** ΔQ =mCΔT ΔS = ΔQ/T

Here T is the average temperature, in kelvins, of the two bodies in which the heat transfer takes place.

Now, the specific heat of water is the amount of heat required to raise the temperature of 1 kg of water by 1^{0} C. From this we can calculate the heat absorbed to raise the temperature of 10 kg of water by 10^{0} C. Therefore,

ΔQ = 4180 × 10 × 10 = 418000 J (By using the first formula).

Now to find the change in entropy, ΔS, use the 2^{nd} formula. First we find the average temperature.

T = (363 + 373)/2 = 736/2 = 368 K

Now put in the formula,

ΔS = 418000/368 = **1135.9 J K**^{-1}