An iron cup of mass 0.180 kg, initially at 18°C is filled with 0.100 kg of hot water initially at 70°C. What is the final temperature after the water and the cup attain thermal equilibrium?
c(iron) = 0.47 J/g°C
I know that you use this equation: mcΔT = mcΔT.
The answer is 61.3°C, but I keep getting a different answer. Any help is appreciated.
c(iron) = 0.47 J/g°C
I know that you use this equation: mcΔT = mcΔT.
The answer is 61.3°C, but I keep getting a different answer. Any help is appreciated.
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The amount of heat that leaves the water must go to the cup:
0.100 kg * 4.18 kJ/( C kg) * (70 - t) = 0.180 kg * 0.47 kJ/(C kg) * (t - 18)
Thus
29.26 - 0.418 t = 0.0846 t - 1.5228
0.5026 t = 30.7828
t = 61.2 C
0.100 kg * 4.18 kJ/( C kg) * (70 - t) = 0.180 kg * 0.47 kJ/(C kg) * (t - 18)
Thus
29.26 - 0.418 t = 0.0846 t - 1.5228
0.5026 t = 30.7828
t = 61.2 C