A hot lump of 26.5 g of copper at an initial temperature of 77.0 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to reach thermal equilibrium.

Question

A hot lump of 26.5 g of copper at an initial temperature of 77.0 °C is placed in 50.0 mL H2O 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 0.385 J/(g·°C)? Assume no heat is lost to surroundings.

Answer 1

The final temperature of the copper and water will be the same and will be denoted by Tf.

Heat lost by the hot copper MUST equal heat gained by the cooler water (conservation of energy)

Use q = mC∆T where q = heat; m = mass; C = specific heat; ∆T = change in temperature

For the hot metal:

q = mC∆T = (26.5 g)(0.385 J/gº)(77º-Tf)

For the water:

q = mC∆T = (50.0 g)(4.184 J/gº)(Tf-25.0º) .. we assumed a density of 1 g / ml for H2O to get the mass

Set the two equal to each other because heat lost by copper must = heat gained by water:

(26.5 g)(0.385 J/gº)(77º-Tf) = (50.0 g)(4.184 J/gº)(Tf-25.0º)

Solve for Tf (final temperature):

785.6 -10.20Tf = 209.2Tf - 5230

219.4Tf = 6015.6

Tf = 27.4ºC

Be sure to check the math