Question

Use the data in the following table to answer the questions below: Specific Heat Capacity Melting Temperature Latent Heat of Fusion Material Aluminium Glass Gold Iron Platinumm Liquid Water Water Ice (J.kg K) 870 840 129 450 126 4180 2110 933 1773 1337 1811 2041 273 273 (kJ.kg) 321 N/A 67.0 209 113 334 334 All quantities are given to at least 3 significant figures Calculate the initial temperature of the horseshoe, giving your final answer in units of Kelvin You should assume that no heat was lost to the environment and that the pot and the water were in thermal equilibrium before the horseshoe entered the water. Recall that the mass of the iron horseshoe was 0.479 kg, and the final temperature of the horseshoe is 29.8 C Answer

A freshly-forged iron horseshoe, with a mass of 0.479 kg is dropped into a 0.200 kg iron pot which contains 1.39 kg of water at 21.8 ^oC.

Answer 1

Assuming Initial temperature of horseshoe is T,

from energy conservation:

Heat gained by water + iron pot = Heat lost by horseshoe

And Q = m*C*dT

Q1 + Q2 = Q3

Mw*Cw*dT1 + Mi*Ci*dT2 = Mh*Ch*dT3

Initial temp of water = 21.8 C

Final temperature of system = 29.8 C, So

dT1 = dT2 = 29.8 - 21.8 = 8.0 C

Initial temp of horseshoe = T = ?

dT3 = T - 29.8

Cw = Specific heat of water = 4180 J/kg-C

Cw = Specific heat of horseshoe and iron pot = 450 J/kg-C

Mass of water = 1.39 kg

Mass of iron pot = 0.200 kg

Mass of horseshoe = 0.479 kg

Using given values:

1.39*4180*8 + 0.200*450*8 = 0.479*450*(T - 29.8)

T = [1.39*4180*8 + 0.200*450*8 + 0.479*450*29.8]/(0.479*450)

T = 248.8 C = 248.8 + 273

T = 521.8 K = 522 K

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