Question

A 1.50-kg iron horseshoe initially at 610°C is dropped into a bucket containing 22.0 kg of water at 26.0°C. What is the final temperature of the water–horseshoe system? Ignore the heat capacity of the container and assume a negligible amount of water boils away.

°C

Answer 1

Assuming final temperature is T of system,

from energy conservation:

Heat gained by water = Heat lost by horseshoe

And Q = m*C*dT

Q1 = Q2

Mw*Cw*dT1 = Mh*Ch*dT2

dT1 = T - 26 & dT2 = 610 - T

Cw = Specific heat of water = 4186 J/kg-C

Cw = Specific heat of horseshoe = 450 J/kg-C

Using given values:

22*4186*(T - 26) = 1.5*450*(610 - T)

T = (1.5*450*610 + 22*4186*26)/(22*4186 + 1.5*450)

Solving above equatipn:

T = 30.25 C

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