Question

3. How many kilograms of water could be heated from a temperature of 31.8°C to 91.3°C with 6220 kJ of heat?

Answer 1

Q = M S ΔT Mass * specific heat water * change in temp

M = Q / (S * ΔT)

ΔT = (91.3 - 31.8) deg C = 59.5 deg C

M = 6220 KJoules / (1.00 kJ / kg deg C * 59.5 deg C)

M = 105 kg