How much heat must a gram of water at 15°C lose to turn into ice at 0°C?
To calculate the amount of heat that must be lost, we need the specific heat capacity and the heat of fusion of water. The specific heat capacity of water is approximately 4.18 J/g°C, and the heat of fusion (or heat required to convert a substance from a solid to liquid or vice versa) is 334 J/g.
First, we need to calculate the heat that the water needs to cool down from 15°C to 0°C:
Q1 = m * c * ΔT
Q1 = 1 g * 4.18 J/g°C * (0°C - 15°C)
Q1 = -62.7 J
Next, we calculate the heat needed for the water to freeze at 0°C:
Q2 = m * ΔHf
Q2 = 1 g * 334 J/g
Q2 = 334 J
The total heat needed for the water to turn into ice is the sum of Q1 and Q2:
Total heat = Q1 + Q2
Total heat = -62.7 J + 334 J
Total heat = 271.3 J
Therefore, a gram of water at 15°C must lose 271.3 J of heat in order to turn into ice at 0°C.
cool the water to 0deg=mass*speheat*15
solifify the water: mass*heatfusion
I get about 95 calories in my head