physics
posted by Julie on .
A hot iron horseshoe (mass = 0.43 kg), just forged (Fig. 1416), 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*(36Ti)+1.60*specheatwater*(2620)=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(T26) = 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!

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