How much heat must be removed from 456 g of water at 25.0°C to change it into ice at -10.0°C? The specific heat of ice is 2090 J/kg ∙ K, the latent heat of fusion of water is 33.5 × 104 J/kg, and the specific heat of water is 4186 J/kg ∙ K.
To find the amount of heat that must be removed from the water to change it into ice, we need to consider two steps: first, cooling the water from 25.0°C to 0°C, and second, freezing the water at 0°C into ice at -10.0°C.
Step 1: Cooling the water from 25.0°C to 0°C
We can use the formula Q = m * c * ΔT, where Q is the heat energy, m is the mass, c is the specific heat, and ΔT is the temperature change.
Q1 = m1 * c1 * ΔT1
Given:
Mass of water (m1) = 456 g = 0.456 kg
Specific heat of water (c1) = 4186 J/kg ∙ K
Initial temperature of water (T1) = 25.0°C
Final temperature we want to reach (T2) = 0°C
First, we need to calculate the temperature change (ΔT1):
ΔT1 = T2 - T1
ΔT1 = 0°C - 25.0°C
ΔT1 = -25.0°C
Now, we can calculate the heat energy (Q1) needed to cool the water:
Q1 = m1 * c1 * ΔT1
Q1 = 0.456 kg * 4186 J/kg ∙ K * (-25.0°C)
Q1 = -47949.6 J (rounded to the nearest tenth)
Step 2: Freezing the water from 0°C to -10.0°C
To freeze the water, we need to remove the latent heat of fusion (Lf), which is given as 33.5 × 10^4 J/kg.
Given:
Mass of water (m2) = 456 g = 0.456 kg
Latent heat of fusion of water (Lf) = 33.5 × 10^4 J/kg
Temperature change (ΔT2) = -10.0°C - 0°C = -10.0°C
Now, we can calculate the heat energy (Q2) needed to freeze the water:
Q2 = m2 * Lf
Q2 = 0.456 kg * (33.5 × 10^4 J/kg)
Q2 = 15276 J
Finally, we can find the total heat energy (Qt) needed to change the water into ice:
Qt = Q1 + Q2
Qt = -47949.6 J + 15276 J
Qt ≈ -32673.6 J (rounded to the nearest tenth)
Therefore, approximately 32673.6 J of heat must be removed from the water at 25.0°C to change it into ice at -10.0°C.