How much heat is needed to convert 6 kg of water at -30 ° to water at 100 °?

Given:

m = 6 kilograms is the mass of water;

T1 = -30 degrees Celsius – initial water temperature;

T2 = 100 degrees Celsius – final water temperature;

c = 4200 J / (kg * C) – specific heat capacity of water.

It is required to determine the amount of energy Q required to heat water from temperature T1 to T2.

Since the condition of the problem is not specified, we assume that the water was in a liquid state even at negative temperatures. Then:

Q = c * m * dT = c * m * (T2 – T1) = 4200 * 6 * (100 + 30) =

= 4200 * 6 * 130 = 25200 * 130 = 3276000 Joules = 3.276 MJ.

Answer: you need to spend energy equal to 3.276 MJ.