A 300.0 g sample of water at 80.0°C is mixed with 300.0 g of water at 10.0°C. Assuming no heat loss to the surroundings, what is the final
temperature of the mixture? The specific heat capacity of liquid water is
4180 J/kg.°C *

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The final temperature of the mixture is 45 °C

Heat lost by warm water = Heat gained by cold water

Qᵥᵥ = Q꜀

MᵥᵥC(Tᵥᵥ – Tₑ) = M꜀C(Tₑ – T꜀)

0.3 × 4180 × (80 – Tₑ) = 0.3 × 4180 × (Tₑ – 10)

1254 (80 – Tₑ) = 1254(Tₑ – 10)

Clear bracket

100320 – 1254Tₑ = 1254Tₑ – 12540

Collect like terms

100320 + 12540 = 1254Tₑ + 1254Tₑ

112860 = 2508Tₑ

Divide both side by 2508

Tₑ = 112860 / 2508

Tₑ = 45 °C

Thus, the final temperature of the mixture is 45 °C

Learn more about heat transfer:

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T_finalmix = 45 [°C].

Explanation:

In order to solve this problem, a thermal balance must be performed, where the heat is transferred from water with the highest temperature, to the water with the lowest temperature, so the final temperature of the mix must be equal once the thermal balance is achieved.

where:

mwater = mass of the water = 300 [g] = 0.3 [kg]

Cp_water = specific heat of the water = 4180 [J/kg*°C]

T_waterinitial = initial temperature of the water = 80 [°C]

T_finalmix = final temperature of the mix [°C]

Now replacing: