An unknown volume of water at 18.2°C is added to 27.0 mL of water at 36.7°C. If the final temperature is 23.5°C, what was the unknown volume? (Assume that no heat is lost to the surroundings; d of water is 1.00 g/mL.)
heat lost by one + heat gained by the other = 0
(mass1 x specific heat x [Tfinal - Tinitil]) + (mass2 x specific heat x [Tfinal - Tinitial]) = 0
You have only one unknown, solve for mass (1 or 2 depending upon how you set up the problem), then since the density of water is assumed to be 1.00 g/mL, the mass will be the same as the volume, in mL. Post your work if you get stuck.
WRONG
please explain more
To solve this problem, we can use the principle of conservation of energy, specifically the equation for heat transfer:
Q1 + Q2 = 0
where Q1 is the heat loss from the hot water and Q2 is the heat gain by the cold water.
We can calculate Q1 and Q2 using the equation:
Q = m * c * ΔT
where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate Q1:
Q1 = m1 * c * ΔT1
where m1 is the mass of the hot water.
Given:
m1 = 27.0 mL (volume)
c = 1.00 g/mL (specific heat of water)
ΔT1 = 36.7°C - 23.5°C (temperature change of the hot water)
Now, let's calculate Q2:
Q2 = m2 * c * ΔT2
where m2 is the mass of the cold water.
Given:
c = 1.00 g/mL (specific heat of water)
ΔT2 = 23.5°C - 18.2°C (temperature change of the cold water)
We know that the heat lost by the hot water is equal to the heat gained by the cold water, so Q1 = -Q2. Therefore:
m1 * c * ΔT1 = -m2 * c * ΔT2
Now, let's substitute the given values into the equation:
(27.0 mL) * (1.00 g/mL) * (36.7°C - 23.5°C) = -(unknown volume) * (1.00 g/mL) * (23.5°C - 18.2°C)
Simplifying the equation:
(27.0 mL) * (13.2°C) = (unknown volume) * (5.3°C)
Dividing both sides by 5.3°C:
(27.0 mL) * (13.2°C) / (5.3°C) = unknown volume
Calculating the unknown volume:
unknown volume = (27.0 mL) * (13.2°C) / (5.3°C)
unknown volume = 67.714 mL
Therefore, the unknown volume of water is approximately 67.714 mL.