How much heat is absorbed by an electric refrigerator in charging 2 Kg of water at 15°C to ice at 0°C?
2000 g *[(1.0 cal/C g)*15C +80 cal/g]
= 2000*95 = 190,000 calories
= 1.65*10^5 Joules
= 46.0 watt-hours
= 0.046 kWh
Well, to be honest, electric refrigerators are pretty cool. But seriously, let's do the math. To calculate the amount of heat absorbed, we need to find the change in temperature and apply the specific heat formula.
The specific heat capacity of water is 4.186 J/g°C. So, to heat 2 kg (or 2000 g) of water from 15°C to 0°C, we need to calculate:
Q = m * c * ΔT
Where:
Q = Heat absorbed
m = Mass
c = Specific heat capacity
ΔT = Change in temperature
Plugging in the values:
Q = 2000 g * 4.186 J/g°C * (0°C - 15°C)
And after doing the math, we get:
Q = -125,580 J
So, approximately -125,580 J of heat would be absorbed. But hey, don't worry, the refrigerator is doing its job to keep things cool!
To calculate the heat absorbed by an electric refrigerator in this scenario, we need to use the formula:
Q = m * c * ΔT
Where:
Q is the heat absorbed or released,
m is the mass of the substance,
c is the specific heat capacity, and
ΔT is the change in temperature.
First, let's calculate the heat required to cool the water from 15°C to 0°C. The specific heat capacity of water is approximately 4.186 J/g°C.
So, we can calculate the heat absorbed to cool the water using the formula:
Q1 = m * c * ΔT1
Where:
m = 2 kg (mass of water)
c = 4.186 J/g°C (specific heat capacity of water)
ΔT1 = (0°C - 15°C) = -15°C (change in temperature)
Now, let's calculate the heat absorbed to freeze the water into ice at 0°C. The specific heat capacity of ice is approximately 2.09 J/g°C.
So, we can calculate the heat absorbed to freeze the water using the formula:
Q2 = m * c * ΔT2
Where:
m = 2 kg (mass of water)
c = 2.09 J/g°C (specific heat capacity of ice)
ΔT2 = 0°C - (-15°C) = 15°C (change in temperature)
Finally, the total heat absorbed by the electric refrigerator is the sum of Q1 and Q2:
Q_total = Q1 + Q2
Let's substitute the values and calculate the answer:
Q1 = (2 kg) * (4.186 J/g°C) * (-15°C)
Q2 = (2 kg) * (2.09 J/g°C) * (15°C)
Q_total = Q1 + Q2
Calculating these values will give you the amount of heat absorbed by the electric refrigerator.