Question

An 8.2 kg, super-cold (-150 degrees C) block of copper is placed in a container with 450 g of 100% liquid water at 0 degree C, and the container is sealed. At what temperature will we find the container much later? (Hint: note that the first thing the water does is to phase change to a solid).

Answer 1

Suppose the final temperature is T.

Now Using energy conservation:

Heat released by water = Heat absorbed by copper

Q1 = Q2

Q1 = Heat released due to phase change + Heat released by ice from 0 C to T C

Q1 = Mw*Lf + Mw*Ci*dT1

Q2 = Heat absorbed by Copper from -150 C to T C

Q2 = Mc*Cc*dT2

dT1 = Tf - Ti = 0 - T = -T

dT2 = Tf - Ti = T - (-150) = T + 150

Mw = mass of water = 450 gm = 0.45 kg

Mc = mass of Copper = 8.2 kg

Cc = Specific Heat capacity of Copper = 390 J/kg-C

Lf = Latent Heat of fusion = 3.34*10^5 J/kg

Ci = Specific Heat of ice = 2090 J/kg-C

Now using given values:

Q1 = Q2

Mw*Lf + Mw*Ci*dT1 = Mc*Cc*dT2

0.45*3.34*10^5 - 0.45*2090*T = 8.2*390*(T + 150)

Now Solving above equation

T = (0.45*3.34*10^5 - 8.2*390*150)/(8.2*390 + 0.45*2090)

T = -79.6 C

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answered by: anonymous