If 0.060kg of ice at 0 ∘C is added to 0.320kg of water at 25∘C in a 0.150-kg aluminum calorimeter cup, what is the final temperature of the water?
To find the final temperature of the water, we can use the principle of conservation of energy. We assume that there is no energy loss to the surroundings.
The total heat gained by the water and aluminum cup should be equal to the total heat lost by the ice to melt and the water to cool down.
Let's calculate the heat gained by the water and aluminum cup first:
Qwater = mwater * cwater * ΔTwater
where:
mwater = mass of water = 0.320 kg
cwater = specific heat capacity of water = 4186 J/kg°C
ΔTwater = change in temperature of water = final temperature - initial temperature = final temperature - 25°C
Qcup = mcup * ccup * ΔTcup
where:
mcup = mass of aluminum cup = 0.150 kg
ccup = specific heat capacity of aluminum = 900 J/kg°C
ΔTcup = change in temperature of aluminum cup = final temperature - initial temperature = final temperature - 25°C
Since the water and aluminum cup are in thermal equilibrium, the total heat gained is equal to the total heat lost:
Qwater + Qcup = Qice
where:
Qice = mice * ΔHfusion + mice * cwater * ΔTfinal_water
mice = mass of ice = 0.060 kg
ΔHfusion = enthalpy of fusion for ice = 333.55 kJ/kg (or 333550 J/kg)
ΔTfinal_water = final temperature - 0°C
Now, let's solve for the final temperature of the water:
Qwater + Qcup = Qice
(mwater * cwater * ΔTwater) + (mcup * ccup * ΔTcup) = (mice * ΔHfusion) + (mice * cwater * ΔTfinal_water)
Substituting the given values:
(0.320 kg * 4186 J/kg°C * (final temperature - 25°C)) + (0.150 kg * 900 J/kg°C * (final temperature - 25°C)) = (0.060 kg * 333550 J/kg) + (0.060 kg * 4186 J/kg°C * ΔTfinal_water)
Simplifying the equation:
(1340.48 * final temperature - 33512) + (135 * final temperature - 3375) = (20013 + 251.16 * ΔTfinal_water)
Combine like terms:
1475.48 * final temperature - 36887 = 20013 + 251.16 * ΔTfinal_water
Rearranging the equation:
1475.48 * ΔTfinal_water - 251.16 * ΔTfinal_water = 36887 - 20013
Simplifying:
1224.32 * ΔTfinal_water = 16874
Dividing by 1224.32:
ΔTfinal_water = 13.78°C
Finally, we can find the final temperature of the water by adding the change in temperature to the initial temperature:
Final temperature = 25°C + 13.78°C
Final temperature = 38.78°C
Therefore, the final temperature of the water is approximately 38.78°C.
To find the final temperature of the system, we can use the principle of conservation of energy. The heat lost by the hot water and the heat gained by the ice must be equal.
First, let's calculate the heat lost by the hot water. We can use the formula:
Q1 = m1 * c1 * ΔT1
where:
m1 is the mass of the hot water (0.320 kg)
c1 is the specific heat capacity of water (4.18 J/g°C or 4180 J/kg°C)
ΔT1 is the change in temperature of the hot water (final temperature - initial temperature)
ΔT1 = Tf - 25°C
Q1 = 0.320 kg * 4180 J/kg°C * (Tf - 25°C)
Next, let's calculate the heat gained by the ice. We can use the formula:
Q2 = m2 * c2 * ΔT2
where:
m2 is the mass of the ice (0.060 kg)
c2 is the specific heat capacity of ice (2.09 J/g°C or 2090 J/kg°C)
ΔT2 is the change in temperature of the ice (final temperature - initial temperature)
ΔT2 = Tf - 0°C
Q2 = 0.060 kg * 2090 J/kg°C * (Tf - 0°C)
According to the principle of conservation of energy, Q1 (heat lost) must be equal to Q2 (heat gained):
Q1 = Q2
0.320 kg * 4180 J/kg°C * (Tf - 25°C) = 0.060 kg * 2090 J/kg°C * (Tf - 0°C)
Simplifying the equation:
(0.320 kg * 4180 J/kg°C * Tf) - (0.320 kg * 4180 J/kg°C * 25°C) = (0.060 kg * 2090 J/kg°C * Tf) - (0.060 kg * 2090 J/kg°C * 0°C)
(1.344 Tf - 10520) = (0.1254 Tf)
1.344 Tf - 0.1254 Tf = 10520
1.2186 Tf = 10520
Tf = 10520 / 1.2186
Tf ≈ 8619.82 °C
Since the final temperature cannot be greater than 100 °C (boiling point of water), the final temperature of the water is approximately 100 °C.