A student mixes 85.7 g of water at 68.0°C with 50.7 g of water at 24.5°C in an insulated flask. What is the final temperature of the combined water

call the 85 g water water1, and the other water2.

sum of heat changes =0
heatfromwater1+heatfromwater1=0
One will be negative, as one of them cools.

mass1*cw*deltaTemp1+mass2*cw+deltaTemp2=0
I assume you can handle it from here. Tf in each will be the same, ti for each is given. A little algebra in this...

To find the final temperature of the combined water, we can use the principle of conservation of energy.

The amount of energy gained or lost by a substance can be calculated using the equation:

q = m * c * ΔT

Where:
- q is the heat energy gained or lost
- m is the mass of the substance
- c is the specific heat capacity of the substance
- ΔT is the change in temperature

Since the two samples of water are mixed in an insulated flask and no energy is gained or lost to the surroundings, the total amount of heat energy gained by the colder water is equal to the total amount of heat energy lost by the hotter water (q_cold = -q_hot).

We can set up the equation and solve for the final temperature (T_final).

q_cold = m_cold * c_water * (T_final - T_cold)
q_hot = m_hot * c_water * (T_final - T_hot)

Since q_cold = -q_hot, we can set them equal to each other:

m_cold * c_water * (T_final - T_cold) = -m_hot * c_water * (T_final - T_hot)

Now we can plug in the given values:
- m_cold = 50.7 g (mass of the cold water)
- T_cold = 24.5°C (initial temperature of the cold water)
- m_hot = 85.7 g (mass of the hot water)
- T_hot = 68.0°C (initial temperature of the hot water)

50.7 g * c_water * (T_final - 24.5°C) = -85.7 g * c_water * (T_final - 68.0°C)

We cancel out c_water from both sides of the equation:

50.7 * (T_final - 24.5) = -85.7 * (T_final - 68.0)

Now we can solve for T_final:

50.7 * T_final - 50.7 * 24.5 = -85.7 * T_final + 85.7 * 68.0

Collecting like terms:

50.7 * T_final + 85.7 * T_final = 85.7 * 68.0 + 50.7 * 24.5

136.4 * T_final = 5789.6 + 1243.35

136.4 * T_final = 7032.95

T_final = 7032.95 / 136.4

T_final ≈ 51.5°C

Therefore, the final temperature of the combined water is approximately 51.5°C.