Question

What is the enthalpy change when `"5 g"` of `Y` are dissolved in `"100 g"` of water to decrease its temperature from `22^@ "C"` to `17^@ "C"`? `C_P = "4.184 J/g"^@ "C"` for pure water near `25^@ "C"`.

Answer 1

`DeltaH = q_P = -"2196.6 J"`.

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It doesn't really matter what `Y` is, since we are only finding `DeltaH` in `"J"`, not `"J/mol"`. We thus don't need the molar mass of `Y`.

At constant pressure, by definition, the heat flow `q_P` is given by

`q_P = DeltaH = mC_PDeltaT`,

where

• `DeltaH` is the change in enthalpy of the solution.
• `m` is the mass of the solution.
• `C_P` is the specific heat capacity of the solution in `"J/g"^@ "C"`.
• `DeltaT` is the change in temperature of the solution (in guess what units? `"K"`? Sure. But we'll use `""^@ "C"`).

As a result, there is not much of an extra step here. It's just regular heat flow calculations, with a different label for what `q` is at constant pressure.

(If it were constant volume, we would have given you an empirically-obtained calorimeter heat capacity in `"kJ/"^@ "C"`, and asked for `DeltaE` instead.)

We assume:

• the same specific heat capacity of water as usual, despite the solution not being pure water
• that `C_P` does not change within the temperature range.

`color(blue)(q_P) = ("5 g + 100 g")("4.184 J/g"^@ "C")(17.0^@ "C" - 22.0^@ "C")`

`=` `color(blue)(-"2196.6 J")`