Question

If 200 g of water is contained in a 300g aluminum calorimeter at 20

Answer 1

E = m * C * T

For the water,

Ec = 200 g * 4186 J/kg/K * (20 + 273) K
Ec = 245.299 kJ

For the aluminum,

Ea = 300 g * 900 J/kg/K * (20+273) K
Ea = 79.11 kJ

For the steam,

Eh = 20 g * 1996 J/kg/K * (100+273) K
Eh = 14.89016 kJ

Now, that total amount of energy is contained in the system.

E = Ec + Ea + Eh
E = 339.299 kJ

Remember, the amount of energy in the system is the same when the hot water is separate from the cold water and aluminum as when they are combined (assuming you don't lose any energy to the environment - in other words, that you have a closed system).

When everything reaches equilibrium, it all has the same temperature, T. So, the combined mass of water has

Ew = 200 g * 4186 J/kg/K * T
Ew = 837.2 J/K * T

The aluminum has

Ea = 300 g * 900 J/kg/K * T
Ea = 270 J/K * T

The steam has

Eh =20*1996*T

Eh = 39.92*T
And finally, since the energy is the same as before, we have

Ea + Ew = E
Ea + Ew + Eh= 339.299 kJ

That's 3 equations and 3 unknowns, so we should be able to solve.

270 J/K * T + 39.92 J/K * T + 837.2 J/K * T = 339.299 kJ

T * (1147.12 J/K) = 339.299 kJ
T = 295.78 K
T = (295.78 - 273) C
T =22.78 C