1 kg of ice at 0° C is mixed with 9 kg of water at 50° C (The latent heat of ice is 3.34x105 J/kg and the specific heat capacity of water is 4160 J/kg). What is the resulting temperature?
Heat into ice = 1*3.34*10^5 +(T-0)(4160)
Heat out of water = 4160 (50-T)(9)
set equal and solve for T
1*3 is one times 3
1 kilogram times latent heat of ice
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To find the resulting temperature after mixing the ice and water, we can use the principle of energy conservation. The heat lost by the water at 50°C is equal to the heat gained by the ice at 0°C.
First, let's calculate the heat lost by the water:
Qwater = mwater * cwater * ΔTwater
Where:
mwater = mass of water = 9 kg
cwater = specific heat capacity of water = 4160 J/kg°C
ΔTwater = change in temperature of water = (resulting temperature - initial temperature)
Now, let's calculate the heat gained by the ice:
Qice = mice * L
Where:
mice = mass of ice = 1 kg
L = latent heat of ice = 3.34x10^5 J/kg
Since the heat lost by the water is equal to the heat gained by the ice, we can equate the two equations:
mwater * cwater * ΔTwater = mice * L
Now, let's solve for ΔTwater:
ΔTwater = (mice * L) / (mwater * cwater)
Substituting the values into the equation:
ΔTwater = (1 kg * 3.34x10^5 J/kg) / (9 kg * 4160 J/kg°C)
Calculating this expression:
ΔTwater = 75.817°C
To find the resulting temperature, we can subtract ΔTwater from the initial temperature of the water:
resulting temperature = 50°C - 75.817°C = -25.817°C
The resulting temperature after mixing the ice and water is approximately -25.817°C.