A 25.0 g block of copper (specific heat capacity 0.380 J/g･°C) at 54.0 °C is placed into 500.0 g of water initially at 20.0 °C.

Question

What is the change in temperature (in °C) of the copper block? (The specific heat capacity of water is 4.184 J/g･°C).

Answer 1

Mass copper = 25.0 g

Specific heat capacity copper = 0.380 J/g degrees C

Temperature Initial copper = 54.0 degrees C

Temperature Final copper = ?

Mass water = 500.0 g

Specific heat capacity water = 4.184 J/g degrees C

Temperature Initial water = 20.0 degrees C

Temperature final water = ?

If both the water and copper are in thermal equilibrium, they will reach the same final temperature. We don’t know either of their final temperatures but we know that they have the same final temperature, Tf. They will also exchange the same amount of heat (as one heats up, the other cools) so we know that their heats q will be the same magnitude, just with one being negative (losing heat) and the other positive.

q lost = - q gained

The copper is losing heat because it’s initially hotter than the water and heat flows from hot to cold. The water will be gaining heat.

heat equation q = c* m * change in T

q lost = q copper

q gained = q water

q copper = (0.380)(25.0)(Tf - 54.0)

q copper = 9.5(Tf - 54.0) = q lost

q water = (4.184)(500.0)(Tf - 20.0)

q water = 2092(Tf - 20.0) = q gained

multiply by -1 to turn this expression negative

-q water = -2092(Tf-20) = - q gained

q lost = - q gained

set the expression for q copper equal to the expression for -q water

9.5(Tf - 54.0) = -2092(Tf-20)

distribute the parentheses on each side

9.5Tf - 9.5(54.0) = -2092Tf -(-2092)(20)

9.5Tf - 513 = -2092Tf + 41840

add 2092Tf to both sides

2101.5Tf - 513 = 41840

add 513 to both sides

2101.5Tf = 42353

divide by 2101.5 on both sides

Tf = 20.2 degrees C

The change in temperature of the copper was

Tf - Ti = 20.2- 54.0 = -33.8 degrees C

The temperature of the copper decreased by 33.8 degrees C

Answer 2

The heat lost by the hot copper must equal the heat gained by the cooler water.

Heat lost by copper = q = mC∆T

q = heat = ?

m = mass = 25.0 g

C = specific heat of copper = 0.380 J/gº

∆T = change in temperature = 54º - Tf where Tf is the final temperature

Heat gained by water = q = mC∆T

q = ?

m = mass = 500.0 g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = Tf - 20.0º

Setting the two heats equal because heat lost must equal heat gained, we have...

(25.0 g)(0.380 J/gº)(54-Tf) = (500 g)(4.184 J/gº)(Tf-20)

513 - 9.5Tf = 2092Tf - 41,840

2080.5Tf = 42,353

Tf = 20.4ºC

(please be sure to check my math)