A hot iron horseshoe (mass = 0.40 kg) which has just beenforged, is dropped into 1.35 L of water in a 0.30-kg iron potintially at 20?C. If the final equilibrium temperature is 25?C,estimate the initial temperature of the hot horseshoe

A hot iron horseshoe (mass = 0.40 kg) which has just beenforged, is dropped into 1.35 L of water in a 0.30-kg iron potintially at 20?C. If the final equilibrium temperature is 25?C,estimate the initial temperature of the hot horseshoe
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Velsenw
Loss of Heat energy = Gian of heat energy by water + Gain ofheat energy by iron pot.
ms*Ss*(tis-tfs) = mw*Sw*(tiw-tfw) + mp*Sp*(tip-tfp)
Where ms = mass of the shoe = 0.40kg.
Ss = specifi heat of shoe = 4400J/kg.K
tis = Intial temperature of shoe.
tfs = Final temperature of shoe = 25o
C. similarly for water and pot.
therefore 0.4*4400*(tis - 25) = 1.35*4186*5+0.3*4400*5
1760*(tis - 25) = 28255.5 + 6600
1760*(tis - 25) = 34855.5
By solving above eqation we get intial temperature of shoe is tis =44.8oC.