An aluminium container of mass 100 gm contains 200gm of ice at -20c heat is added to system at rate of 100 cal/s find temp of system after 4minutes

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

97.3 °C 

Why the rest of the question?