A heater cansupply 98000J an hour. If a lot of ice is already at 0 degrees, how much ice can be turned to water?
check my work please: Q/l=m
98000J/kg / 3.35 * 10^5J = 029kg of ice can be turned to water
0.29 kg I think you mean
yes, 0.29kg is what I meant, is it correct?
To determine how much ice can be turned into water using a heater that supplies 98,000 J per hour, you can use the equation:
Q / l = m
Where:
Q = total heat supplied by the heater (in joules)
l = heat of fusion for ice (in joules per kilogram)
m = mass of ice that can be turned into water (in kilograms)
Since the ice is already at 0 degrees Celsius, and the heat of fusion for ice is 3.35 x 10^5 J/kg, you can plug in these values into the equation:
Q = 98,000 J/hr
l = 3.35 x 10^5 J/kg
Substituting these values into the equation:
98,000 J/hr / (3.35 x 10^5 J/kg) = m
Now, let's calculate the value:
98,000 J/hr divided by 3.35 x 10^5 J/kg equals approximately 0.29 kg of ice that can be turned into water.
So, according to your calculations, approximately 0.29 kg or 290 grams of ice can be turned into water in one hour using a heater that supplies 98,000 J/hr.