a refrigerator can convert 400g of water at 20 degree celcius to ice at-10°c in 3hours.find the average rate of heatextraction from the water in joulesper seconds
Answers
ANSWER :-
- Average rate of heat extraction is 16.33 Joules/sec or 16.33Watt.
GIVEN :-
- Mass of water = 400g = 0.4kg
- Water is converted from 20°C to -10°C
- Time taken = 3hours
TO KNOW :-
Specific heat capacity of water,
ㅤㅤㅤㅤㅤㅤㅤ★ C(water) = 4200/kg/°C
Specific heat capacity of ice ,
ㅤㅤㅤㅤㅤㅤㅤ★ C(ice) = 2100/kg/°C
Specific Latent heat of fusion of ice ,
ㅤㅤㅤㅤㅤㅤㅤ★ L(ice) = 336000/kg
FORMULA USED :-
ㅤㅤㅤㅤㅤㅤㅤ♦ Q = mC∆T
ㅤㅤㅤㅤㅤㅤㅤㅤ♦ Q = mL
Here ,
- Q = Heat extraction
- m = mass
- C = specific heat
- ∆T = Initial Temperature - Final Temperature
- L = Specific Latent heat
HOW TO SOLVE :-
→ Water is initially at 20°C . It is converted to 0°C(water) . We will find Heat extracted in this case . Let the Heat Extraction be Q1.
→ Then , water is converted from 0°C(water) to 0°C(ice) . We will find Heat extracted in this case .Let the Heat Extraction be Q2.
→ Then , water is converted from 0°(ice) to -10°C .
We will find Heat extracted in this case . Let the Heat Extracted be Q3.
Now , Total heat extraction (Q) is the sum of heat extracted in all 3 cases.
Q = Q1 + Q2 + Q3
Rate of heat extraction is the Heat extraction per unit time.
Rate of heat extraction = Q/t
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Refer the attachment for calculations.
