Question

Question #7cc47

Answer 1

Here's what I got.

Explanation:

The idea here is that the heat given off by the sample of ethanol must account for heating the water present inside the calorimeter and the calorimeter itself.

`color(blue)(ul(color(black)(q_"ethanol" = - (q_"water" + q_"calorimeter"))))" " " "color(darkorange)((1))`

The minus sign is used here because heat given off carries a negative sign

The thing to remember about performing a reaction in a bomb calorimeter is that volume is always kept constant, i.e. `DeltaV = 0`, which implies that no work is done, i.e. `W = 0`.

Therefore, you can say that

`color(blue)(ul(color(black)(DeltaU = q)))" " " "color(darkorange)((2))`

The change in the internal energy of the system is equal to the heat supplied to the system

So, the first thing to do here is to figure out the heat needed to increase the temperature of `"3000 g"` of water by

`DeltaT = 26.225^@"C" - 24.284^@"C" = 1.941^@"C"`

Water has a specific heat of

`c_"water" = "4.18 J g"^(-1)""^@"C"^(-1)`

which means that it takes `"4.18 J"` of heat to increase the temperature of `1^@"C"` of water by `1^@"C"`.

You know that

`color(blue)(ul(color(black)(q = m * c * DeltaT)))`

Here

• `q` is the heat lost or gained by the substance
• `m` is the mass of the sample
• `c` is the specific heat of the substance
• `DeltaT` is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

In this case, you will have

`q_"water" = 3000 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1))))color(red)(cancel(color(black)(""^@"C"^(-1)))) * 1.941color(red)(cancel(color(black)(""^@"C")))`

`q_"water" = "24,340 J"`

Expressed in kilojoules, this will be

`q_"water" = "24.34 kJ"`

Next, calculate the heat needed to increase the temperature of the calorimeter by `1.941^@"C"`. You know that you need `"2.71 kJ"` of heat to increase the temperature of the calorimeter by `1^@"C"`, which means that you will have

`1.941 color(red)(cancel(color(black)(""^@"C"))) * "2.71 kJ"/(1color(red)(cancel(color(black)(""^@"C")))) = "5.26 kJ"`

The heat needed to increase the temperature of the water and of the calorimeter will thus be equal to

`"24.34 kJ " + " 5.26 kJ" = "29.6 kJ"`

Therefore, you can equation `color(darkorange)((1))` to say that the reaction produced

`q_"ethanol" = - "29.6 kJ"`

This means that when `"1.000 g"` of ethanol undergo combustion, `"29.6 kJ"` of heat are being given off.

You can now calculate the change in internal energy by using equation `color(darkorange)((2))`

`color(darkgreen)(ul(color(black)(DeltaU = - "29.6 kJ g"^(-1))))`

Alternatively, you can convert this to kilojoules per mole of ethanol by using the compound's molar mass

`1.000 color(red)(cancel(color(black)("g"))) * "1 mole ethanol"/(46.07color(red)(cancel(color(black)("g")))) = "0.021706 moles ethanol"`

You will thus have

`1 color(red)(cancel(color(black)("mole ethanol"))) * (-"29.6 kJ")/(0.021706color(red)(cancel(color(black)("moles ethanol")))) = -"1364 kJ"`

Therefore,

`color(darkgreen)(ul(color(black)(DeltaU = - "1360 kJ mol"^(-1))))`

I'll leave both answers rounded to three sig figs, but keep in mind that you only have one significant figure for the mass of water.