Question

An insulated container is used to hold 43.6 g of water at 20.6 °C. A sample of copper weighing 11.0 g is placed in a dry test tube and heated for 30 minutes in a boiling water bath at 100.0°C. The heated test tube is carefully removed from the water bath with laboratory tongs and inclined so that the copper slides into the water in the insulated container. Given that the specific heat of solid copper is 0.385 J/(g·°C), calculate the maximum temperature of the water in the insulated container after the copper metal is added.

Answer 1

m(water) = 43.6 g

T(water) = 20.6 oC

C(water) = 4.184 J/goC

m(copper) = 11.0 g

T(copper) = 100.0 oC

C(copper) = 0.385 J/goC

T = 23.9 oC

We will be using heat conservation equation

Let the final temperature be T oC

use:

heat lost by copper = heat gained by water

m(copper)*C(copper)*(T(copper)-T) = m(water)*C(water)*(T-T(water))

11.0*0.385*(100.0-T) = 43.6*4.184*(T-20.6)

4.235*(100.0-T) = 182.4224*(T-20.6)

423.5 - 4.235*T = 182.4224*T - 3757.9014

T= 22.4 oC

Answer: 22.4 oC