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Suppose you mix 100.0 g of water at 22.8 oC with 75.0 g of water at 76.3  oC. What will be the final temperature of the mixed water, in oC?

Question

Suppose you mix 100.0 g of water at 22.8 oC with 75.0 g of water at 76.3  oC. What will be the final temperature of the mixed water, in oC?

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Solution

To solve this problem, we need to use the concept of heat transfer, which states that in an isolated system (like our mixture of water), heat lost will be equal to heat gained.

We can express this mathematically as:

m1c(Tf-T1) = -m2c(Tf-T2)

where: m1 = mass of the first substance (100.0 g) T1 = initial temperature of the first substance (22.8 °C) m2 = mass of the second substance (75.0 g) T2 = initial temperature of the second substance (76.3 °C) Tf = final temperature of the mixture c = specific heat capacity of water (4.18 J/g°C)

We can plug in the known values and solve for Tf:

100.0 g * 4.18 J/g°C * (Tf - 22.8 °C) = -75.0 g * 4.18 J/g°C * (Tf - 76.3 °C)

Solving this equation will give us the final temperature of the mixture.

This problem has been solved

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