A technician needs to determine the amount of energy required to increase the temp. of a vat containing 12 kg of water. The water enters the vat from the water mains at a temp. of 15 deg C. The technician knows that the final temp. required is
90 deg C. How much energy is required?
Show work please.
change in temp = 90 - 15 = 75 deg C
heat in = mass * specific heat * change in temp
= 12 kg * specific heat * 75 deg C
I do not know specific heat in Joules or calories or kilocalories or whatever you are using in your class / kilogram deg C
To determine the amount of energy required to increase the temperature of the water in the vat, we can use the specific heat formula. The specific heat capacity of water is approximately 4.186 J/g°C.
First, we need to calculate the change in temperature of the water. The water enters the vat at 15°C and needs to reach a final temperature of 90°C.
ΔT (change in temperature) = final temperature - initial temperature
ΔT = 90°C - 15°C
ΔT = 75°C
Next, we need to calculate the mass of the water in the vat in grams, as the specific heat capacity given is in J/g°C. We can convert the 12 kg to grams by multiplying it by 1000 since 1 kg is equal to 1000 g.
Mass = 12 kg * 1000 g/kg
Mass = 12,000 g
Now we can calculate the amount of energy required using the formula:
Energy = mass * specific heat capacity * ΔT
Energy = 12,000 g * 4.186 J/g°C * 75°C
Energy = 3,156,720 J
Therefore, approximately 3,156,720 Joules of energy is required to increase the temperature of the vat containing 12 kg of water from 15°C to 90°C.