Question

A 130.0 g piece of copper (specific heat 0.38 J/g･°C) is heated and then placed into 400.0 g of water initially at 20.7°C. The water increases in temperature to 22.2°C. What is the initial temperature of the copper? (The specific heat of water is 4.18 J/g･°C and the density of water is 1.00 g/mL).

Answer 1

Make aheat balance

heat lost= heat gained

-Qcopper = Qwater

recall that this is snesible heat so

Q = m*C*(Tf-Ti)

therefore, apply for each

Qcopper = m copper * Cp copper * (Tf-Tcopper)

Qwater = m water * Cp water * (Tf- Twater)

substitutein balance:

-m copper * Cp copper * (Tf-Tcopper) = m water * Cp water * (Tf- Twater)

substitute data

-m copper * Cp copper * (Tf-Tcopper) = m water * Cp water * (Tf- Twater)

-130 * 0.38 * (22.2-Tcopper) = 400 * 4.18 * (22.2 - 20.7)

(22.2-Tcopper)=  (400 * 4.18 * (22.2 - 20.7) ) / (-130 * 0.38)

Tcopper = -( -50.76-22.2) = 72.96° C

Tcopper initially = 72.96°C