Question

1. (a) A 300.0 g copper kettle contains 1.000 kg of water at 20.0 °C. How much energy must you supply to heat the kettle + water to 98.6 °C? (b) How much propane must you burn to provide this energy, if the transfer of the heat of combustion is 35 %?

`C_text(Cu) = "0.385 J·°C"^"-1"·"g"^"-1"`

For propane, `Δ_text(c)H = "2202.0 kJ·mol"^"-1"`

Answer 1

Warning! Long Answer. Here's the answer to Question 1.

Explanation:

a) Energy required

The heat energy `q` needed to heat an object is given by the formula

`color(blue)(bar(ul(|color(white)(a/a)q = mcΔTcolor(white)(a/a)|)))" "`

where

`m color(white)(l)=` the mass of the object
`c color(white)(m)=` the specific heat capacity of the object and
`ΔT` = the change in temperature

You have two objects to be heated: the copper kettle and the water.

∴ `q = q_text(Cu) + q_text(water) = q_1 + q_2 = m_1c_1ΔT + m_2c_2ΔT`

In this problem,

`m_1 = "300.0 g"`
`c_1color(white)(ll) = "0.385 J·°C·g"^"-1"`

`ΔT = "(98.6 – 20.0) °C = 78.6 °C"`

`m_2 = "1000 g"`
`c_2color(white)(l) = "4.184 J·°C·g"^"-1"`

∴ `q =`

`300.0 color(red)(cancel(color(black)("g"))) × "0.385 J"·color(red)(cancel(color(black)("°C·g"^"-1"))) × 78.6 color(red)(cancel(color(black)("°C"))) + 1000 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("°C·g"^"-1"))) × 78.6 color(red)(cancel(color(black)("°C"))) = "9078 J + 328 900 J" = "338 000 J" = "338 kJ"`

The energy required is 338 kJ.

b) Mass of propane required

The heat `q` obtained from combustion of a substance is given by the formula

`color(blue)(bar(ul(|color(white)(a/a)q = nΔ_text(c)Hcolor(white)(a/a)|)))" "`

where

`ncolor(white)(mll) =` the number of moles of the substance and
`Δ_text(c)H =` its molar heat of combustion

You need to supply 338 kJ of energy to the kettle, but only 35 % of the enthalpy of combustion is actually transferred.

∴ `"Energy supplied" = 338 color(red)(cancel(color(black)("kJ used"))) × "100 kJ supplied"/(35 color(red)(cancel(color(black)("kJ used")))) = "966 kJ supplied"`

The enthalpy of combustion of propane is -2202.0 kJ/mol.

∴ `n = q/(Δ_text(c)H) = 966 color(red)(cancel(color(black)("kJ")))× ("1 mol C"_3"H"_8)/(2202.0 color(red)(cancel(color(black)("kJ")))) = "0.439 mol C"_3"H"_8`

`"Mass of C"_3"H"_8 = 0.439 color(red)(cancel(color(black)("mol C"_3"H"_8))) × (44.10 "g C"_3"H"_8)/( 1 color(red)(cancel(color(black)("mol C"_3"H"_8)))) =19 color(white)(l) "g C"_3"H"_8`