1) Compute the amount of heat needed to raise the temperature of 1 kg of water f
ID: 1483715 • Letter: 1
Question
1) Compute the amount of heat needed to raise the temperature of 1 kg of water from its freezing pint to its normal boiling point. (in Joules)
2) A 10kg. lead brick is dropped from the top of a 600-m tall tower and falls to ground. Assuming all of its energy goes to heat the brick, what is the temperature increase (in °C)? Note: Specific heat capacity of Lead is 130 J/kg-°C
3) What is the efficiency of an engine if it takes in 15000 J of chemical potential in energy and outputs 5000 J of work in the same time increment?
4) 1st Law of Thermodynamics states that the change in internal energy of substance equals the_____?______ done on it plus the heat transferred to it. The 2nd Law of Thermodynamics states that no device can be built that will extract heat from the source and deliver work or energy without ejecting some_______?_______ to a lower-temperature reservoir. Therefore, the entropy of the universe always___________?__________ (increases/decreases).
Explanation / Answer
Solution:
Part (1)
Mass of water m = 1 kg,
Initial temperature of water T = 0oC
Final temperature of water T’ =100oC
Specific heat of water, c = 4180 J/kg.oC
Amount of heat Q needed to raise the temperature from 0oC to 100oC can be found by,
Q = c*m*(T’ – T)
Q = (4180 J/kg.oC)*(1 kg)*( 100oC - 0 oC)
Q = 418000 J
Hence the answer is 418000 J.
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Part (2)
Solution:
Mass of the lead brick, m = 10 kg
Height from which it is dropped, h = 600 m
Specific heat capacity of the lead c = 130 J/kg. oC
The change in temperature T = T’ – T = to be found.
The gravitational potential energy of the lead brick at the height of h = 600 m is given by,
V = m*g*h
This potential energy is converted into the heat Q, hence the tempearure of the lead brick increases,
Q = V
c*m*T = m*g*h
c*T = g*h
(130 J/kg. oC)*T = (9.81m/s2)*(600m)
T = (9.81m/s2)*(600m)/ (130 J/kg. oC)
T = 45.277 oC
Thus the temperature increase is 45.3 oC
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Part (3)
Energy fed to engine, Q = 15000 J
output from the enegine, W = 5000 J.
The efficiency of the engine is given by,
= W/Q
= 5000 J / 15000 J
= 0.3333
Thus the answer is 0.33
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Part (4)
Solution:
From the definition of first law of thermodynamics, it follows that,
Eint = Q – W
Where Eint is the change internal energy, Q is the heat supplied and W is workdone by on it. Thus
1st Law of Thermodynamics states that the change in internal energy of substance equals the WORK done on it plus the heat transferred to it.
The 2nd Law of Thermodynamics states that no device can be built that will extract heat from the source and deliver work or energy without ejecting some HEAT to a lower-temperature reservoir. Therefore, the entropy of the universe always INCREASES.
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