Question

Resources Lx Give Up? A hot lump of 27.7 g of copper at an initial temperature of 67.8 °C is placed in 50.0 mL H,O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the copper and water, given that the specific heat of copper is 0.385J/(g.°C)? Assume no heat is lost to surroundings. Trinal = TOOLS *10'

Answer 1

The expressions of heat released by copper and that absorbed by water are matched:

m * cp * ΔT water = - m * cp * ΔT copper

50 * 4.18 * (Tf - 25) = - 27.7 * 0.385 * (Tf - 67.8)

Clears Tf = 27.1 ° C

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