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March 27, 2015

March 27, 2015

Posted by **Julie** on Monday, April 23, 2012 at 2:35pm.

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 -
**Elena**, Monday, April 23, 2012 at 3:31pmhorseshoe 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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