Given:
m = 52 kg. the mass of water
ΔT = 22 °C, the temperature increase
Note that
c = 4.184 J/(g-°C), the specific heat of water.
The quantity of heat required is
Q = mcΔT
= (52,000 g)*(4.184 J/(g-°C))*(22 °C)
= 4.7865 x 10⁴ J
Because 1 J = 2.39 x 10⁻⁴ cal. the heat required is
Q = (4.7865 x 10⁴ J)*(2.39 x 10⁻⁴ cal/J) = 1143.97 cal ≈ 1144 cal
Answer: 1144 cal
