Question

Another chemistry question over thermodynamics?

Answer 1

`T_text(f) = "71 °C"`

Explanation:

There are two heat transfers to consider:

`"heat released by reaction + heat absorbed by water = 0"`
`color(white)(mmmmmm)q_1 color(white)(mmmmml)+color(white)(mmmmmm) q_2 color(white)(mmmml)= 0`
`color(white)(mmmmm)nΔ _text(r)H color(white)(mmmml)+ color(white)(mmmm)mC_text(s)ΔT color(white)(mmml)= 0`

Let's calculate each of these heats separately.

Step 1. Calculate `q_1`

`q_1 = nΔ_text(r)H = 0.110 color(red)(cancel(color(black)("mol"))) × ("-131 kJ"·color(red)(cancel(color(black)("mol"^"-1")))) = "-14.41 kJ" = "-14 410 J"`

Step 2. Calculate `q_2`

`m color(white)(l)= "75 g"`
`C_text(s) color(white)(l)= "4.184 J·°C"^"-1""g"^"-1"`
`ΔT = ?`

∴ `q_2 = 75 color(red)(cancel(color(black)("g"))) × "4.184 J·°C"^"-1"color(red)(cancel(color(black)("g"^"-1"))) × ΔT = 314 color(white)(l)ΔTcolor(white)(l) "J·°C"^"-1"`

Step 3. Calculate `T_text(f)`

`ΔT = T_text(f) - T_text(i)`

`q_1 + q_2 = 0`

`"-14 410" color(red)(cancel(color(black)("J"))) + 314 color(white)(l)ΔTcolor(red)(cancel(color(black)("J")))·"°C"^"-1" = 0`

`ΔT = "14 410"/("314 °C"^"-1") = "45.9 °C"`

`T_text(f) = T_text(i) + ΔT = "(25 + 45.9) °C = 71 °C"`