SOLVED PROBLEMS #1
1. When 2,100 J is removed from 0.10 kg of a substance, its temperature is observed
to decrease from 40oC to 10oC. What is the specific heat of the substance?
Solution: Using the formula: Q = mcΔT where Q = 2,100 J
2,100 J = 0.10 kg (c) (40 – 10)oC
c = 700 J/kg-oC
2. How much heat is required to raise an 800 gm copper pan from 15oC to 90oC if the
pan contains 1.0 kg of water?
Solution: Since both the pan and the water absorb heat, the total heat required is:
Q = Heat absorbed by the pan + Heat absorbed by the water
= 0.8 kg (386 J/kg-K) (90 – 15) + 1.0 kg (4185 J/kg-K) (90 – 15)
= 337,000J
3. In a calorimetry experiment, 0.50 kg of a metal at 100oC is added to 0.50 L of
water at room temperature in an aluminum calorimeter cup. The cup has a mass of
250 gm. If the final temperature of the mixture is 25oC, what is the specific heat of
the metal?
Solution:
In this problem, the calorimer and the water in it will absorb the heat released
by the hot metal.
0.50 kg (c) (100 – 25)oC = 0.5 kg (4185) (25 – 20) + 0.250 (920) (25 – 20)
c = 309.67 J/ kg-oC
4. Determine the amount of heat a freezer has to remove from 1.5 kg of water at 22oC
to make ice at – 10oC.
Solution:
a) Heat required to change the temperature of water at 22oC to 0oC
Q1 = mcΔT = (1.5 kg) (4186 J/kg-oC)(22 – 0)oC
= 138.14 J
b) Heat required to change water at 0oC to ice at 0oC
Q2 = mLf = (1.5 kg)(3.33 x 105 J/kg)
= 4.995 x 105 J
c) Heat to be removed to lower ice from 0oC to – 10oC
Q3 = mcΔT = (1.5 kg) (2100 J/kg-oC) [0 – (–10)oC]
= 3.15 x 104 J
Total Heat required: Qtotal = 138.14 J + 4.995 x 105 J + 3.15 x 104 J
Qtotal = 5.31 x 105 J
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