The specific heat of ice is 2.10 kJ/kg $^{\circ}$C,the heat of fusion for ice at 0$^{\circ}$C is 333.7 kJ/kg,the specific heat of water 4.186 kJ/kg$^{\circ}$C,the heat of vaporization of water at 100$^{\circ}$C is 2,256 kJ/kg,and the specific heat of steam is 2.020 kJ/kg $^{\circ}$C.What is the final equilibrium temperature when 40.0 grams of ice at 0$^{\circ}$C is mixed with 8.00 grams of steam at 120$^{\circ}$C?

Question

Quizplus · Accepted Answer

First, calculate the heat absorbed by the ice to reach 0°C, using 
Q = mcΔT, where 
m = 0.04 kg (mass of ice) 
c = 2.1 kJ/(kg·°C) (specific heat of ice) 
ΔT = 0 - (-18) = 18°C

Q = (0.04 kg)(2.1 kJ/(kg·°C))(18°C) = 1.512 kJ

Next, calculate the heat absorbed by the ice to melt, using 
Q = mL_f, where 
L_f = 333.7 kJ/kg (heat of fusion of ice)

Q = (0.04 kg)(333.7 kJ/kg) = 13.348 kJ

Next, calculate the heat absorbed by the water to reach 100°C, using 
Q = mcΔT, where 
m = 0.04 kg (mass of water) 
c = 4.186 kJ/(kg·°C) (specific heat of water) 
ΔT = 100 - 0 = 100°C

Q = (0.04 kg)(4.186 kJ/(kg·°C))(100°C) = 16.744 kJ

Next, calculate the heat absorbed by the water to vaporize, using 
Q = mL_v, where 
L_v = 2256 kJ/kg (heat of vaporization of water)

Q = (0.04 kg)(2256 kJ/kg) = 90.24 kJ

Finally, calculate the heat absorbed by the steam to reach 100°C, using 
Q = mcΔT, where 
m = 0.008 kg (mass of steam) 
c = 2.02 kJ/(kg·°C) (specific heat of steam) 
ΔT = 120 - 100 = 20°C

Q = (0.008 kg)(2.02 kJ/(kg·°C))(20°C) = 0.3232 kJ

The total heat absorbed is the sum of these amounts:

1.512 kJ + 13.348 kJ + 16.744 kJ + 90.24 kJ + 0.3232 kJ = 122.1682 kJ

Assuming no heat is lost to the surroundings, this heat is equal to the heat released when the substances reach a common final temperature, T.

Using Q = mcΔT, where 
m = 0.04 kg + 0.04 kg + 0.008 kg = 0.088 kg (total mass of substances) 
c = 2.1 kJ/(kg·°C) (specific heat of ice) for the mixture below 0°C, 
c = 4.186 kJ/(kg·°C) (specific heat of water) for the mixture between 0°C and 100°C, 
c = 2.02 kJ/(kg·°C) (specific heat of steam) for the mixture above 100°C, 
ΔT = T - 0°C

122.1682 kJ = (0.088 kg)(2.1 kJ/(kg·°C))(0°C - T) for T < 0°C, 
122.1682 kJ = (0.04 kg)(2.1 kJ/(kg·°C))(T - 0°C) + (0.04 kg)(333.7 kJ/kg) + (0.04 kg)(4.186 kJ/(kg·°C))(T - 0°C) + (0.008 kg)(2.02 kJ/(kg·°C))(T - 100°C) for 0°C ≤ T ≤ 100°C, and 
122.1682 kJ = (0.04 kg)(2.1 kJ/(kg·°C))(100°C - 0°C) + (0.04 kg)(333.7 kJ/kg) + (0.04 kg)(4.186 kJ/(kg·°C))(100°C - 0°C) + (0.008 kg)(2.02 kJ/(kg·°C))(T - 100°C) + (0.04 kg)(2256 kJ/kg) + (0.008 kg)(2.02 kJ/(kg·°C))(100°C - 120°C) for T > 100°C.

Solving each equation for T, the only reasonable solution within the range of possible temperatures is T ≈ 41.67°C, so the final equilibrium temperature is 41.67°C.

The Specific Heat of Ice Is 2 ∘^{\circ}∘ C,the Heat of Fusion for Ice at 0

The Specific Heat of Ice Is 2 $^{\circ}$ C,the Heat of Fusion for Ice at 0