How much ice at zero degrees celsius is needed to cool 250g of water at 25 degrees celsius to zero degrees celsius.
.250 kg * specific heat of water * 25 = heat of fusion of water * mass of ice in kg
To solve this problem, you need to calculate the amount of heat energy required to cool the water from 25°C to 0°C and then determine how much ice will be needed to absorb this amount of heat.
First, you need to calculate the heat energy (Q) required to cool the water from 25°C to 0°C using the formula:
Q = m * c * ΔT
Where:
Q is the heat energy in Joules
m is the mass of the water in grams
c is the specific heat capacity of water (approximately 4.18 Joules/gram°C)
ΔT is the change in temperature in degrees Celsius
Given:
m = 250g
c = 4.18 J/g°C
ΔT = 25°C - 0°C = 25°C
Substituting these values into the formula:
Q = 250g * 4.18 J/g°C * 25°C
Q ≈ 26,125 Joules
Next, you need to determine the amount of heat energy (Q') that can be absorbed by a given amount of ice during its phase change from solid to liquid (i.e., melting). The heat of fusion (or latent heat of fusion) for ice is approximately 334 J/g.
Q' = Q / (heat of fusion of ice)
Q' = 26,125 J / 334 J/g
Q' ≈ 78.04 g
Therefore, approximately 78.04 grams of ice at 0°C would be needed to cool 250 grams of water from 25°C to 0°C.