Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 3 (Q.No. 19)
19.
Latent heat absorbed by 1 lb of water at 212°F, when it is changed to steam at 212°F, may be around __________ BTU.
Discussion:
10 comments Page 1 of 1.
Ramsajanpaswan said:
2 years ago
As per my knowledge, the right answer is 970BTU.
(4)
Rushi said:
8 years ago
It's latent heat i.e. phase change it should be more than 180 BTU.
Answer may be 970.
Answer may be 970.
(2)
Rajesh Singh said:
4 years ago
970 BTU is the right answer.
(2)
Rajesh Singh said:
4 years ago
The Right answer is 972 BTU.
(2)
TANMAY SHAH said:
2 years ago
I think the Right answer is Option-B.
(2)
Dheeraj Verma said:
2 years ago
I agree, The right answer is 970.
(2)
Pankaj said:
5 years ago
970 is the correct answer.
(1)
Smriti Baroi said:
1 year ago
970 btu would be the right answer.
(1)
Philipdarwin said:
8 years ago
I agree @Rushi.
Example - Boiling Water at 212°F and 0psig.
At atmospheric pressure - 0psig - water boils at 212°F. 180 Btu/lb of energy is required to heat 1 lb of water to saturation temperature 212°F.
Therefore, at 0 psig and 212°F - the specific enthalpy of water is 180 Btu/lb.
Another 970 Btu/lb of energy is required to evaporate the 1 lb of water at 212°F to steam at 212 °F. Therefore, at 0 psig - the specific enthalpy of evaporation is 970 Btu/lb.
The total specific enthalpy of the steam (or heat required to evaporate water to steam) at atmospheric pressure and 212°F can be summarized to;
hs = (180 Btu/lb) + ( 970 Btu/lb).
= 1150 Btu/lb.
Example - Boiling Water at 212°F and 0psig.
At atmospheric pressure - 0psig - water boils at 212°F. 180 Btu/lb of energy is required to heat 1 lb of water to saturation temperature 212°F.
Therefore, at 0 psig and 212°F - the specific enthalpy of water is 180 Btu/lb.
Another 970 Btu/lb of energy is required to evaporate the 1 lb of water at 212°F to steam at 212 °F. Therefore, at 0 psig - the specific enthalpy of evaporation is 970 Btu/lb.
The total specific enthalpy of the steam (or heat required to evaporate water to steam) at atmospheric pressure and 212°F can be summarized to;
hs = (180 Btu/lb) + ( 970 Btu/lb).
= 1150 Btu/lb.
Maqsood said:
6 years ago
Please provide the clear explanation for the answer.
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