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.
.456 [ 4186(25-0) + 335,000 + 2090(0+10) ]
To calculate the total heat that must be removed, we need to consider three steps:
1. Heating the water from 25.0°C to 0°C.
2. Melting the ice at 0°C.
3. Cooling the melted ice from 0°C to -10.0°C.
Step 1: Heating the water from 25.0°C to 0°C.
The specific heat of water is 4186 J/kg ∙ K, and the initial temperature is 25.0°C. We need to find the heat required to raise the temperature of 456 g of water to 0°C.
Q1 = m * c1 * ΔT1
Q1 = (0.456 kg) * (4186 J/kg ∙ K) * (0 - 25.0°C)
Q1 = -4815 J
Step 2: Melting the ice at 0°C.
The latent heat of fusion of water is 33.5 × 10^4 J/kg. We need to find the heat required to melt 456 g of ice.
Q2 = m * L2
Q2 = (0.456 kg) * (33.5 × 10^4 J/kg)
Q2 = 15276 J
Step 3: Cooling the melted ice from 0°C to -10.0°C.
The specific heat of ice is 2090 J/kg ∙ K, and the temperature change is from 0°C to -10.0°C. Since the ice is already melted, we need to consider only the specific heat, not the latent heat.
Q3 = m * c3 * ΔT3
Q3 = (0.456 kg) * (2090 J/kg ∙ K) * (-10.0 - 0°C)
Q3 = -9549 J
The total heat that must be removed is the sum of Q1, Q2, and Q3:
Total heat = Q1 + Q2 + Q3
Total heat = -4815 J + 15276 J - 9549 J
Total heat = 917 J
Therefore, 917 J of heat must be removed from 456 g of water at 25.0°C to change it into ice at -10.0°C.
To calculate the amount of heat that must be removed from the water to change it into ice, we need to consider two processes: cooling the water from 25.0°C to 0°C, and then changing it from 0°C to -10.0°C as ice.
First, let's find the heat required to cool the water from 25.0°C to 0°C using the specific heat formula:
Q1 = m * c * ΔT
Where
Q1 is the heat required,
m is the mass of water,
c is the specific heat of water, and
ΔT is the change in temperature.
Given:
mass of water (m) = 456 g
specific heat of water (c) = 4186 J/kg ∙ K
initial temperature (T₁) = 25.0°C
final temperature (T₂) = 0°C
Converting the mass to kg:
mass (m) = 456 g = 0.456 kg
Calculating Q1:
Q1 = 0.456 kg * 4186 J/kg ∙ K * (0°C - 25.0°C)
Q1 = -47828.8 J
So, the heat required to cool the water from 25.0°C to 0°C is -47828.8 J.
Next, let's find the heat required to change the water at 0°C to ice at -10.0°C. This process involves the latent heat of fusion:
Q2 = m * Lf
Where
Q2 is the heat required,
m is the mass of water, and
Lf is the latent heat of fusion.
Given:
mass of water (m) = 456 g
latent heat of fusion (Lf) = 33.5 × 10^4 J/kg
Converting the mass to kg:
mass (m) = 456 g = 0.456 kg
Calculating Q2:
Q2 = 0.456 kg * 33.5 × 10^4 J/kg
Q2 = 15276 J
So, the heat required to change the water at 0°C to ice at -10.0°C is 15276 J.
To find the total heat required to change 456 g of water at 25.0°C to ice at -10.0°C, we need to add the heat required for cooling and the heat required for phase change:
Total heat = Q1 + Q2
Total heat = -47828.8 J + 15276 J
Total heat = -32552.8 J
Therefore, approximately 32,552.8 J of heat must be removed from 456 g of water at 25.0°C to change it into ice at -10.0°C.