How much heat is required to bring 100 g of water at a temperature of 10 ° C to a boil?

Given: m (the mass of water that needs to be brought to a boil) = 100 g (0.1 kg); t (the temperature that the water had before the heat transfer) = 10 ºС.

Constants: С (specific heat capacity of water) = 4200 J / (kg * ºС); tк (temperature at which boiling starts) = 100 ºС.

The required amount of heat to bring water to a boil: Q = C * m * (tc – t) = 4200 * 0.1 * (100 – 10) = 4200 * 0.1 * 90 = 37800 J (37.8 kJ).

Answer: Requires 37.8 kJ of heat.