practice problem 1
practice problem 2
This is a conservation of energy problem. The heat gained by the ice will be equal to the heat lost by the coffee.
+Qice = −Qcoffee
This mixing problem is more complicated than the ones in the previous section, however. The ice must first warm up to its melting point (a temperature change), then it has to melt (a phase change), and then the liquid has to warm up (another temperature change). The coffee has less to do. It just has to cool down.
|[mc∆T] cold ice|
|+||[mc∆T] melted ice|
|= −||[mc∆T] coffee|
The final mixture will end up at one temperature. (Watch the order of subtraction when dealing with temperature changes.)
|mice(2,090 J/kg C°)(0 − −18.5 °C)|
|+||mice(4,200 J/kg C°)(55 − 0 °C)|
|= −||(0.355 kg)(4,200 J/kg C°)(55 − 85 °C)|
|mice(603,665 J/kg)||= (44,730 J)|
|mice||= 0.0741 kg|
This is about three ice cubes.
(74.1 g)/(23.5 g) ≈ 3 ice cubes
practice problem 3
practice problem 4