Specific Heat Capacity Show work Please!

Question

A water heater warms 35-L (35 kg) of water from a temperature of 22.7 ∞C to a temperature of83.7∞C. Determine the amount of energy (in kilojoules) required.

[8933 kJ] <=== Answer

Answer 1

We use the equation q = mC∆T

q = heat = ?

m = mass of water = 35 kg

C = specific heat of water = 4.184 kJ/kgº

∆T = change in temperature = 83.7º - 22.7º = 61.0º

q = (35 kg)(4.184 kJ/kgº)(61º)

q = 8933 kJ = 8900 kJ (2 sig. figs.)