Question

A 95.0 g sample of copper (cmc001-1.jpg = 0.20 J/°C·g) is heated to 82.4°C and then placed in a container of water (cmc001-2.jpg = 4.18 J/°C·g) at 22.0°C. The final temperature of the water and the copper is 25.1°C. What was the mass of the water in the original container? Assume that all heat lost by the copper is gained by the water. Use the formulas below to help in your problem-solving.

-qmc001-3.jpg = qmc001-4.jpg
-cmc001-5.jpgmmc001-6.jpgmc001-7.jpgTmc001-8.jpg = cmc001-9.jpgmmc001-10.jpgmc001-11.jpgTmc001-12.jpg

0.246 g Hmc001-13.jpgO

4.73 g Hmc001-14.jpgO

84.0 g Hmc001-15.jpgO

36,700 g Hmc001-16.jpgO

Answer 1

Use heat lost by the copper = heat gained by the water.

m_c c_c (T_c,i − T_f) = m_w c_w (T_f − T_w,i)

Plug in values:
95.0 g × 0.20 J/(g·°C) × (82.4 − 25.1)°C = m_w × 4.18 J/(g·°C) × (25.1 − 22.0)°C

Left side: 95.0×0.20×57.3 = 1088.7 J
Right side factor for m_w: 4.18×3.1 = 12.958 J/g

m_w = 1088.7 / 12.958 ≈ 84.0 g

Answer: 84.0 g of water.