Question

What is the final temperature of a system if 11.00 g of gold at 90.0°C is placed in 11.00 g of water at 29.00 °C? The molar heat capacity of gold is 25.41 J/(mol·°C) and the heat capacity of water is 4.18 J/(g·°C). °C

Answer 1

heat lost by gold = heat gained by water

q_gold = q_gain

mC_goldT = mC_waterT

11.00 x 0.130 (90 -X) = 11 x 4.18 (X - 29)

1.43 (90-X) = 45.98 (X - 29)

128.7 - 1.43 X = 45.98X - 1333.42

47.41 X = 1462.12

X = 30.84⁰C