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A hot iron horseshoe (mass = 0.43 kg), just forged (Fig. 14-16), is dropped into 1.60 L of water in a 0.26 kg iron pot initially at 20°C. If the final equilibrium temperature is 26°C, estimate the initial temperature of the hot horseshoe.

Could someone please help me out with how to do this. Thank you!



The sum of the heats gained is zero.

heatgainedbyIron+heatgainedbyWater=0
.43*specheatiron*(36-Ti)+1.60*specheatwater*(26-20)=0
solve for Ti


T is final iron temp
a liter of water is a kilogram
heat out of horseshoe = heat into water + heat into pot
.43 Ciron(T-26) = 1.6 Cwater (6 ) + .26 Ciron(6)

I get 1348.26 degrees in Celsius which does not make any sense to me and its wrong, so could someone please help. Thank you!

  • physics -

    horseshoe m1 = 0.43 kg
    water m2 = 1.6•10^-3•1000 = 1.6 kg
    iron pot m3 = 0.26 kg

    specific heats
    water = 4186 J/kg•C
    iron = 450 J/kg•C

    1.6• 4186• (26 - 20) + 0.25• 450•(26 - 20) = 0.43•450•(t- 26)

    t = 237 oC

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