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

Question

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

1 Answer

Impact of this question

2131 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-07T06:15:00

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

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.

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

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

Science

Math

Humanities

... and beyond

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

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is the enthalpy change when "5 g"5 g"5 g" of YYY are dissolved in "100 g"100 g"100 g" of water to decrease its temperature from 22^@ "C"22∘C22^@ "C" to 17^@ "C"17∘C17^@ "C"? C_P = "4.184 J/g"^@ "C"CP=4.184 J/g∘CC_P = "4.184 J/g"^@ "C" for pure water near 25^@ "C"25∘C25^@ "C".

1 Answer

Related questions

Impact of this question

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