Question

A copper pot with a mass of 0.480 kg contains 0.170 kg of water, and both are at a temperature of 20.0 ?C . A 0.270 kg block of iron at 83.0 ?C is dropped into the pot.

Find the final temperature of the system, assuming no heat loss to the surroundings.

Answer 1

Let T (deg C) be the equilibrium temp of system

Heat lost by iron block = gained by copper & water

mi*ci (83 - T) = mc*cc*(T -20) + mw*cw*(T -20)

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m=mass, c-specific heat capacity

c-copper, w-water and i-iron

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0.27*0.45*10^3 (83 - T) =

=0.48*0.385*10^3*(T -20) + 0.17*4.18*10^3*(T -20)

121.5 (83 - T) = 184.8*(T -20) + 710.6*(T -20)

121.5 (83 - T) = 895.4*(T -20)

1016.9 T = 27992.5

T = 27.53 deg C