Question

14.44) A copper pot with a mass of 0.510 kg contains 0.195 kg of water, and both are at a temperature of 21.5 ∘C . A 0.275 kg block of iron at 84.5 ∘C is dropped into the pot.

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

Answer 1

Let T (⁰C) be the equilibrium temp of system

Heat lost by iron block = gained by copper & water
mi*ci (84.5 - T) = mc*cc*(T -21.5) + mw*cw*(T -21.5)
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m=mass, c-specific heat capacity
c-copper, w-water and i-iron
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0.257*0.45*10^3 (84.5 - T) = 0.5*0.385*10^3*(T -21.5) + 0.195*4.18*10^3*(T -21.5)

112.5 (84.5 - T) = 192.5*(T -21.5) + 710.5*(T -21.5)
112.5 (84.5 - T) = 903*(T -21.5)
1015.5 T = 28920.75
T = 28.48⁰ C