1.To get a feeling for the amount of energy needed to heat water, recall from Table 11-1 that the kinetic energy of a compact car moving at 100 km/h is 2.9 105 J. What volume of water (in liters) would 2.9 105 J of energy warm from room temperature (17°C) to boiling (100°C)?
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2.A 2.55 102 W electric immersion heater is used to heat a cup of water. The cup is made of glass and its mass is 3.00 102 g. It contains 235 g of water at 15°C. How much time is needed to bring the water to the boiling point? Assume that the temperature of the cup is the same as the temperature of the water at all times and that no heat is lost to the air.
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3.A 2.85 102 kg cast-iron car engine contains water as a coolant. Suppose the engine's temperature is 35.0°C when it is shut off. The air temperature is 10.0°C. The heat given off by the engine and water in it as they cool to air temperature is 5.0 106 J. What mass of water is used to cool the engine?
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Answers & Comments
Verified answer
Your use of decimal points is rather confusing, but I'll do my best.
1.
Energy required = mass x specific heat capacity of water x change in temperature
E=mcDT
E/cDT=m
105/4.2x83=0.301 kg or 0.3 litres.
Not sure about the units on this one.
2.
E=mcDT
E=235 x 4.2 x 85 = 83895 Joules to heat the water
E=102 x 0.84 x 85 = 7282.8 Joules to heat the glass.
Total energy required = 83895 + 7282.8 = 91177.8 J
Power = E/t
W=J/s
s=J/W
s = 91177.8/102 = 893.9 seconds
3.
This one is a really confusing question - I reiterate, make them clearer. From what I can gather,
E=mcDT (water) mcDT (engine)
(E - mcDT(e))/cDT(w) = m(w)
(106 - (102 x 0.45 x 25))/(4.2 x 25) = 11.9 kg
Hope that helps, although it is not all that brilliant. Took me a while anyway :)