A sample of water containing 2.00 moles is initially at 30.0°C. If the sample absorbs 2.00 kJ of heat, what is the final temperature of the water? (specific heat of water = 4.184 J g^-1 °C^-1)

Question

Quizplus · Accepted Answer

The heat absorbed (2.00 kJ = 2000 J) is used to increase the temperature of the water. Using q = mcΔT, where m = 2.00 moles × 18.015 g/mol = 36.03 g, c = 4.184 J/g°C, and q = 2000 J, solve for ΔT: ΔT = q/(mc) = 2000 J / (36.03 g × 4.184 J/g°C) ≈ 13.3°C. The final temperature is 30.0°C + 13.3°C = 43.3°C.

A Sample of Water Containing 2