1) First calculate the total heat added to the system:
time 4 min * 60 s / min= 240 s
Heat = rate * time = 100 cal/s * 240 s = 24,000 cal
2) Calculate the change due to each element of the systemr:
2.1 Aluminun container, Q Al = mass * Specific heat * ΔT =
mass = 100 g
Specific heat (from a table) = 0.22 cal / (g °C)
ΔT = T - (-20°C)
Q Al = 100 g * 0.22 cal /(g°C) * (T + 20) = 22(T + 20)
2.2 Q ice from -20 °C to 0°C
mass = 200 g
Specific heat = 0.49 cal / (g°C)
ΔT = (0°C - (-20°C)) = 20°C
2.3 Q ice =200g * 0.49 cal / (g°C) * 20°C = 1,960 cal
2.4 Latent fussion heat of ice
From a table: 80 cal / g
Q Lf = 200 g * 80 cal/(g) = 16,000 cal
3) Heating of the water from 0°C to final T
Specific heat = 1 cal / (g°C)
Q w = 200g * 1 cal / (g°C) * (T - 0) = 200T
5) Final calculus
24,000 cal = Q Al + Q ice + Q Lf + Qw =
= 22(T + 20) + 1960 cal + 16,000 cal + 200T = 22T + 440 + 1960 + 200T =
24,000 = 222T + 2400 => 222T = 24000 - 2400 = 21,600 =>
T = 21,600 / 222 = 97.3 °C
Answer: 97.3 °C
