Question

9. Suppose you place a 12 kg piece of Titanium (c=540 J/kgK) at a temperature of 62 C on top of a 0.8 kg block of ice at a temperature of -7 C. The ice heats and eventually melts. Find the final temperature of the water.

Please help me figure out how to get the answer

Answer: 12.58 C

Answer 1

Suppose equilibrium temperature is T, then

Heat lost by Titanium = Heat gained by ice

Q1 = Q2

Q1 = Mt*Ct*dT

Mt = mass of titanium = 12 kg

Ct = 540 J/kg-K

dT = 62 - T

Q2 = energy needed from -7 C to 0 C + Energy to melt the ice + energy needed to reach the final temperature T

Q2 = Mi*Ci*dT1 + Mi*Lf + Mi*Cw*dT2

Mi = mass of ice = 0.8 kg

Ci = specific heat of ice = 2090 J/kg-K

dT1 = 0 - (-7) = 7

dT2 = T - 0 = T

Lf = latent heat of fusion = 3.34*10^5 J/kg

Cw = specific heat of water = 4186 J/kg-K

So,

Mt*Ct*dT = Mi*Ci*dT1 + Mi*Lf + Mi*Cw*dT2

12*540*(62 - T) = 0.8*2090*7 + 0.8*3.34*10^5 + 0.8*4186*T

T*(12*540 + 0.8*4186)= 12*540*62 - (0.8*2090*7 + 0.8*3.34*10^5)

T = [12*540*62 - (0.8*2090*7 + 0.8*3.34*10^5)]/(12*540 + 0.8*4186)

T = 12.50 C

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