Question

Question #99f0f

Answer 1

The temperature change will be +9 °C.

There are two heats to consider:

`"heat released by combustion + heat absorbed by water" = 0`

`q_1 + q_2 = 0`

`nΔ _text(r)H+ mcΔT = 0`

In this problem,

`n = 3.2 color(red)(cancel(color(black)("g propane"))) × "1 mol propane"/(44.10 color(red)(cancel(color(black)("g propane")))) = "0.0726 mol propane"`

`Δ_text(r)H = "-2220 kJ/mol"`

`m = "mass of water" = "4000 g"`

`c = "4.184 J°C"^"-1""g"^"-1"`.

`q_1 = 0.0726 "mol" × "-2 200 kJ"/(1 "mol") = "-160 kJ"`

`q_2 = 4000 cancel("g") × "4.184 J°C"^"-1"cancel("g"^"-1") × ΔT = "17 000 J°C"^"-1" × ΔT = "17 kJ°C"^"-1" × ΔT`

`"-160" color(red)(cancel(color(black)("kJ"))) + 17 color(red)(cancel(color(black)("kJ")))"°C"^"-1" × ΔT = 0`

`ΔT = 160/("17 °C"^"-1") = "9 °C"`