If you combine 270.0 ml of water at 25.00 °c and 100.0 ml of water at 95.0 °c, what is the final temperature of the mixture? use 1.00 g/ml as the density of water.

43.92 °C.

Explanation:

If we combine warm water with cold one, the heat will transfer from the hot water to the cold until reaching an equilibrium temperature.

The amount of heat released by hot water = the amount of heat absorbed by cold water.

The amount of heat released to or absorbed by water (Q) = m. c.ΔT.

where, Q is the amount of heat released to or absorbed by water.

m is the mass of water.

c is the specific heat capacity of water (c = 4.186 J/g.°C).

ΔT is the temperature difference (final T - initial T).

Since, d of water = 1.00 g/mL.

the mass of cold water at 25.0 °C = d. V = (1.00 g/mL) (270.0 mL) = 270.0 g.

the mass of warm water at 95.0 °C = d. V = (1.00 g/mL) (100.0 mL) = 100.0 g.

Q released by warm water = Q absorbed by cold water.

- (m. c.ΔT) released by warm water = (m. c.ΔT) absorbed by cold water.

Since c is the same so,

- (m.ΔT) released by warm water = (m.ΔT) absorbed by cold water.

- [ (100 g) (final T - 95.0 °C) ] = (270 g) (final T - 25.0 °C).

- 100 final T + 9500 = 270 final T - 6750.

270 final T + 100 final T = 9500 + 6750

370 final T = 16250.

∴ final T = 16250/370 = 43.92 °C.