If a 0.345 kg mass of iron at 75 degrees Celsius is placed in a 1 Liter container of water at 4.00 degrees Celsius, what will the final temperature of the two be when they reach equilibrium?

Answer 1

0.345(kg) Iron @ 75(°C) → 1(L) H2O @ 4.00(°C)

Find final temp(°C) at equilibrium

Q = mass(m) × specific heat(c) × change of temp(ΔT) or Q = m × c × ΔT

m_Iron = 0.345(kg) = 345(g) // m_H2O = 1(L) × 1(kg/L) (density of H2O) = 1(kg) = 1000(g)

c_Iron = 0.45(J/g×°C) // c_H2O = 4.18(J/g×°C)

T1_Iron = 75(°C) // T2_H2O = 4.00(°C) // T2 = Final Temp(°C)

Q_lost = Q_gained at the equilibrium

Q_lost = Q_Iron = m × c × (T1 - T2) = 345(g) × 0.45(J/g×°C) × (75(°C) - T2)

Q_gained = Q_H2O = m × c × (T2 - T1) = 1000(g) × 4.18(J/g×°C) × (T2 - 4.00(°C))

Q_lost = Q_gained

345(g) × 0.45(J/g×°C) × (75(°C) - T2) = 1000(g) × 4.18(J/g×°C) × (T2 - 4.00(°C))

155.25(J/°C) × (75(°C) - T2) = 4180(J/°C) × (T2 - 4.00(°C))

11643.75(J) - 155.25(J/°C) × (T2) = 4180(J/°C) × (T2) - 16720(J)

28363.75(J) = 4335.25(J/°C) × (T2)

6.54(°C) = T2 (Final Temp)