Find BTU Required for Water Cooling ?

AI Thread Summary
To calculate the BTU/hr required to cool 30 liters of water from 50°C to 25°C, the specific heat of water (4.184 J/g°C) is used along with the mass of the water (30,000 grams). The temperature change is 25°C, leading to a heat removal calculation. The discussion highlights confusion regarding unit conversions and the relationship between liters, kilograms, and pounds. The goal is to determine the compressor capacity needed to achieve this cooling within 10 to 15 minutes.
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Homework Statement


[/B]
Calculate the BTU/hr required for water to lower its temperature to find the compressor capacity required

- Cool 30 Liters of water from 50 degree Celsius(T1) to 25 Degree Celsius(T2) – ( 122 to 77 Fahrenheit )
-Water in a closed vessel, Static, not flowing.
-Max Time Available to cool the water - 10 to 15 mins (not sure if time is relevant for finding the BTU, But it will be ideal to find the compressor which can do the job in the above time)

Homework Equations



Specific heat equation
Heat added = specific heat x mass x (tfinal - tinitial)

The Attempt at a Solution


[/B]
Specific heat of water = 4.184
Mass = 30 litres = 30,000 grams(approx)
Tfinal - Tinitial = 25 Degree C

bit confused with the units used so I guess I am getting wrong answer
Kindly help to fix thisThanks
 
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Google "British thermal unit."
 
1 liter = 1kg = 2.205 lbm

1 C = 1.8 F

Heat capacity of water = ##1 \frac{BTU}{lbm\ F}##
 
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