How much heat is required to increase the temperature of my 24 feet round 4 1/3

Question

How much heat is required to increase the temperature of my 24 feet round 4 1/3 feet deep pool by 2.0 degrees Celsius from 18 to 20 degrees Celsius? The total volume of pool is unknown. Please calculate volume of pool in liters, than solve for required heat. . I posted this question earlier and the answer was given in inches not sure if it matters? Plase show math and explanation for equation.

Given that Diameter D= 24 ft, R = 12 ft, Height H = 52 inch

Therefore volume V = pi*R2* H = 3.14*122* 52 =23512.32 inch3 = 395.297 litre..........Ans.

We know the heat = mcdt

Mass m = density*volume = 1 Kg/Litre * 395.297 litre = 395.297 Kg

specific heat of water =4186 J/Kg.K

so Heat = 395.297*4186*(20-18) = 33,09,426.48 Joule............Ans.

Explanation / Answer

Given that

Diameter(D) =24ft

Then the radius (R) =12ft =144inch

We know that 1ft=12inch

Then height of the pool (H) =4 (1/3) =13/3 =4.33ft =4.33*12inch =52inch

V = pi*R2* H = 3.14*122* 52 =23512.32 inch3 = 395.297 litre

We know that 1cubic inch =0.0163871lit

We also know that

Heat =mcdt

Mass m = density*volume = 1 Kg/Litre * 395.297 litre = 395.297 Kg

specific heat of water =4186 J/Kg.K

so Heat = 395.297*4186*(20-18) = 33,09,426.48 J =3.309*106J

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