11. A 648-g empty iron kettle is put on a stove. How much heat, in joules, must
ID: 1514747 • Letter: 1
Question
11. A 648-g empty iron kettle is put on a stove. How much heat, in joules, must it absorb to raise its temperature from 15.0°C to 37.0°C? (The specific heat for iron is 113 cal/kg•C°, 1 cal = 4.190 J)
6740 J
11,300 J
1610 J
16,100 J
12. Heat is added to a pure substance in a closed container at a constant rate. The figure shows a graph of the temperature of the substance as a function of time. If Lf = latent heat of fusion and Lv = latent
heat of vaporization, what is the value of the ratio Lv / Lf for this substance?
5.0
4.5
7.2
3.5
1.5
13. If we use 67 W of power to heat 148 g of water, how long will it take to raise the temperature of the water from 15°C to 25°C? The specific heat of water is 4190 J/kg • K.
93 s
5.3 s
22 s
114 h
Explanation / Answer
11)
Heat = m*c*(deltaT)
Heat = 0.648 kg x 113 cal/kg x 22 = 1610.928 J
12)
The lengths of the two flat parts of the graph will tell you what you need.
The length of the higher one ( 7 units ) represents the amount of heat added during vaporization
and
The length of the lower one ( 2 units ) represents the amount of heat added during melting (the latent heat of fusion). Divide the first length by the second.
The value of the ratio Lv / Lf for this substance = 7/2 = 3.5
13)
Well, first let's get the heat transfer equation out:
q = mcT,
where q is the heat (theoretically work),
m is the mass in grams,
c is the specific heat,
and T is the change in temperature.
Calculate the heat first.
q = mcT
q = (148 g)(4.1813 J/g C)(10 C)
q = 6192.76 J
You would need 6192.76 J of heat to heat the water up from 15 C to 25 C. Now, plug this into the work equation (because the heat can be classified as work in the form of chemical energy):
P = W/t
where P is the power,
W is the work,
and t is the time (what you are solving for).
P = W/t
67 W = (6192.76 J)/t
t = 92.42925373 s = 93 secs (approx)
It would take 103.213 seconds to heat the water up with perfect efficiency.
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