Question

0.3 kg of steam at 100°C is added to 1.91 kg of ice at 0°C. Determine the temperature of the mixture once thermal equilibrium is reached.

Answer 1

Using energy conservation:

Suppose the final temperature is T, then

Heat gained by ice = Heat released by steam

Q1 + Q2 =  Q3 + Q4

Q1 = Mi*Lf = energy required to change ice into water

Q2 = Mi*Cw*dT2 = energy required to change 0 C water into T C water

Q3 = Ms*Lv = energy required to change steam into water

Q4 = Ms*Cw*dT4 = energy required to change 100 C water into T C water

Mi*Lf + Mi*Cw*dT2 = Ms*Lv + Ms*Cw*dT4

Using given values:

1.91*3.337*10^5 + 1.91*4186*(T - 0) = 0.3*2.256*10^6 + 0.3*4186*(100 - T)]

637367 + 7995.26*T = 676800+ 125580 - 1255.80*T

T = (676800 + 125580 - 637367)/(1255.80 + 7995.26)

T = 17.84 C

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