An aluminum vessel weighing 0.5 kg contains 0.2 kg of zinc at a temperature of 500 ° C. How much heat will be released

An aluminum vessel weighing 0.5 kg contains 0.2 kg of zinc at a temperature of 500 ° C. How much heat will be released when the vessel with zinc is cooled to 20 ° C? How much alcohol needs to be burned to get the same amount of heat?

M (al) = 0.5 kg
M (zn) = 0.2 kg
dT = 500 – 20 = 480 °
Q =?
m (alcohol) =?
Q = Qal + Qzn
Qal = m * c * dT = 0.5 * 880 * 480 = 211200 J
Qzn = 0.2 * 380 * 480 = 36480 J
Q = 211200 + 36480 = 247680 J
q specific heat of combustion.
q (alcohol) 2.7 * 10 ^ 7 J / kg
Q = m * q m = Q / q = 247680 / 2.7 * 10 ^ 7 = 91.73g