Sensible Heat
Practice
practice problem 1
solution
This is a problem about comparing the energy needed for some task (heating a pot of water)…
Qout = mc∆T Qout = (1.75 kg)(4200 J/kg °C)(80 − 20 °C) Qout = 441,000 J |
to the energy consumed by the device responsible for getting the job done (electric energy)…
Win = Pt Win = (850 W)(10 min)(60 s/min) Win = 510,000 J |
The ratio of these two quantities is the efficiency of the device…
η = Qout/Win η = (441,000 J)/(510,000 J) η = 86.5% |
practice problem 2
solution
This is a classic problem in conservation of energy. The heat lost by the hot object (in this case, the aluminum cube) equals the heat gained by the cold object (in this case, the water bath). Note how the order of subtraction affects the calculation of ∆T. We'll use the specific heat of water at 20 °C since that's basically the temperature of the experiment.
−Qhot = | +Qcold |
−(mc∆T)aluminum = | +(mc∆T)water |
(0.100 kg) |
(2.00 kg) |
Tfinal = | 20.8 °C |
These results are quite sensible (no pun intended). The temperature of the water, which is more massive and has a higher specific heat, has barely changed at all.
practice problem 3
solution
Answer it.
practice problem 4
solution
Answer it.