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

43.92 °C.

Explanation:

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.

answerspecific gravity;

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.