To solve this problem, we can use the principle of heat transfer known as the heat equation:
heat gained by cold substance = heat lost by hot substance
The heat gained by the cold water can be calculated by the equation:
Qcold = mcΔT
Where:
Qcold is the heat gained by the cold water (unknown volume),
m is the mass of the cold water,
c is the specific heat capacity of water (which is approximately 4.18 J/g·°C),
ΔT is the change in temperature of the cold water.
The heat lost by the hot water can be calculated using the same equation:
Qhot = mcΔT
Where:
Qhot is the heat lost by the hot water (24.4 mL),
m is the mass of the hot water,
c is the specific heat capacity of water (which is approximately 4.18 J/g·°C),
ΔT is the change in temperature of the hot water.
Since no heat is lost to the surroundings, the heat gained by the cold water must be equal to the heat lost by the hot water.
So, we can set up an equation:
mcoldΔTcold = mhotΔThot
Now, we'll plug in the values we know:
mhot = 24.4 mL * 1.00 g/mL (density of water is 1.00 g/mL) = 24.4 g
ΔThot = 35.0°C - 23.5°C = 11.5°C
mcold = unknown volume * 1.00 g/mL (density of water is 1.00 g/mL)
ΔTcold = 23.5°C - 18.2°C = 5.3°C
After substituting these values into the equation, we can solve for the unknown volume:
(unknown volume * 1.00 g/mL) * 5.3°C = 24.4 g * 11.5°C
To isolate the unknown volume, we'll divide both sides by 5.3 g/°C:
unknown volume * 1.00 g/mL = 24.4 g * 11.5°C / 5.3 g/°C
Now, we'll solve for the unknown volume:
unknown volume = (24.4 g * 11.5°C / 5.3 g/°C) / (1.00 g/mL)
Calculating this expression will give you the unknown volume of water that was added.