Question

A 0.440 kg iron horseshoe that is initially at 555°C is dropped into a bucket containing 18.1 kg of water at 24.0°C. What is the final equilibrium temperature (in °C)? Neglect any heat transfer to or from the surroundings. Do not enter units.

Answer #1

mass of iron M = 0.44 kg

Initial temprature of the iron t = 555 ^{o}C

Mass of water m = 18.1 kg

Initial temprature of water t ' = 24 ^{o}C

Specific heat of iron C = 450 J / kg ^{o}C

Specific heat of water c = 4186 J / kg ^{o}C

Heat lost by iron = Heat gain by water

MC(t - T ) = mc(T-t ')

199.8 (555-T ) = 75766.6 (T-24)

555-T = 379.21 (T-24)

= 379.21 T - 9101

380.21 T =9656

T = 9656 /380.21

= 25.39 ^{o}C

The final equilibrium temperature T= 25.39

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