Q=55034175 J
c(ice)=2060 J/kg•℃
λ=330000 J/kg
c(water)=4183 J/kg•℃
Q=Q₁+Q₂+Q₃
Q=c(ice)m(12.15-0)+λm+c(water)m(t-0),
Q/m=12.15•c(ice) +λ+c(water)•t.
Solve for “t”
I know my formulas for sensible and latent heat but don't know how to approach this.
c(ice)=2060 J/kg•℃
λ=330000 J/kg
c(water)=4183 J/kg•℃
Q=Q₁+Q₂+Q₃
Q=c(ice)m(12.15-0)+λm+c(water)m(t-0),
Q/m=12.15•c(ice) +λ+c(water)•t.
Solve for “t”
First, let's calculate the sensible heat required to raise the temperature of the ice from -12.15°C to 0°C.
The formula for sensible heat is:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case:
m = 150 kg (mass of the ice)
c = 2.09 kJ/kg°C (specific heat capacity of ice)
ΔT = 0°C - (-12.15°C) = 12.15°C
Plugging the values into the formula:
Q = 150 kg * 2.09 kJ/kg°C * 12.15°C = 3724.92 kJ
So, 3724.92 kJ of heat is required to raise the temperature of the ice to 0°C.
Next, we need to determine the amount of heat needed to convert the ice at 0°C to water at 0°C. This is known as latent heat.
The latent heat formula is:
Q = m * L
Where:
Q is the heat transferred
m is the mass of the substance
L is the latent heat of fusion
In this case:
m = 150 kg (mass of the ice)
L = 333.55 kJ/kg (latent heat of fusion of ice)
Plugging the values into the formula:
Q = 150 kg * 333.55 kJ/kg = 50032.5 kJ
So, 50032.5 kJ of heat is required to convert the ice at 0°C to water at 0°C.
Now, we can add up the sensible heat and latent heat to get the total heat transferred:
Total heat transferred = Sensible heat + Latent heat
Total heat transferred = 3724.92 kJ + 50032.5 kJ = 53757.42 kJ
Finally, we need to calculate the temperature of the resulting water.
Q = m * c * ΔT
In this case:
m = 150 kg (mass of the water)
c = 4.18 kJ/kg°C (specific heat capacity of water)
ΔT = ?
Rearranging the formula:
ΔT = Q / (m * c)
ΔT = 53757.42 kJ / (150 kg * 4.18 kJ/kg°C) = 81.770°C
Therefore, the temperature of the resulting water is 81.770°C.