Answer to Question #88337 in General Chemistry for devin

Question

A 61.26 gram sample of iron (with a specific heat of 0.450 J/goC) is heated to 100.0 ºC. It is then transferred to a coffee cup calorimeter containing 49.61 g of water (specific heat of 4.184 J/ g ºC) initially at 20.63 ºC. If the final temperature of the system is 23.59, what was the heat absorbed (q) of the calorimeter? (Hint: remember that the total heat absorbed by the water AND the calorimeter = the heat released by the iron.) Record your answer as a whole number (assume the sign is positive).

Answer 1

Q_Fe = Q_H2O + Q_cal.

Q_Fe = C_Fe × m_Fe × (T_2Fe – T_1Fe).

Q_H2O = C_H2O × m_H2O × (T_2H2O – T_1H2O).

0.450 J/g^oC × 61.26 g × (100.0 ^oC – 23.59 ^oC) = 4.184 J/g^oC × 49.61 g × (23.59 ^oC – 20.63 ^oC) + Q_cal.

2106.3945 J = 614.4020 J + Q_cal.

Q_cal = 2106.3945 J – 614.4020 J = 1491.9925 J ≈ 1492 J.