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.

Question

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!

Answer 1

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