Question

How long will it take a 3 kW electric heater with an 80 % efficiency to heat a 25 kg copper pot containing 120 kg of water from 5 °C to 65 °C? How much will it cost if the cost of electricity is 6.3 ¢/kWh?

The specific heat capacity of water is `"4.186 J·K"^"-1""g"^"-1"`.
The specific heat capacity of copper is `"0.385 J·K"^"-1""g"^"-1"`.

Answer 1

(a) It will take `color(blue)("3.56 h")` and (b) it will cost `color(blue)(67 ¢)`.

Explanation:

(a) Calculate the time.

There are two heats involved.

`"Total heat" = "heat absorbed by water" + "heat absorbed by copper"`

`q_"tot" = q_"w" + q_"Cu"`

`q_"tot" = m_"w"c_"w"ΔT + m_"Cu"c_"Cu"ΔT`

`c_"w" = "4.186 J·K"^(-1)"g"^(-1) = "4.186 kJ·K"^(-1)"kg"^(-1)`

`c_"Cu" = "0.385 J·K"^(-1)"g"^(-1) = "0.385 kJ·K"^(-1)"kg"^(-1)`

`ΔT = "(65-5) °C"= "60 °C" = "60 K"`

`q_"tot" = 120 color(red)(cancel(color(black)("kg"))) × "4.186 kJ·"color(red)(cancel(color(black)("K"^(-1)"kg"^(-1)))) × 60 color(red)(cancel(color(black)("K"))) + 25 color(red)(cancel(color(black)("kg"))) ×"385 kJ·"color(red)(cancel(color(black)("K"^(-1)"kg"^(-1)))) × 60 color(red)(cancel(color(black)("K")))`

`q_"tot" = "30 139 kJ + 577.5 kJ" = "30 717 kJ"`

Your heater must supply more energy because it is only 80 % efficient.

It must generate 100 kJ to supply 80 kJ.

∴ `"Energy generated" = "30 717" color(red)(cancel(color(black)("kJ needed"))) × "100 kg generated"/(80 color(red)(cancel(color(black)("kg needed")))) = "38 396 kJ"`

`"3 kW" = "3 kJ·s"^-1`

∴ `"Time" = "38 396" color(red)(cancel(color(black)("kJ"))) × "1 s"/(3 color(red)(cancel(color(black)("kJ")))) = "12 799 s"`

`12 799 color(red)(cancel(color(black)("s"))) × (1 color(red)(cancel(color(black)("min"))))/(60 color(red)(cancel(color(black)("s")))) × "1 h"/(60 color(red)(cancel(color(black)("min")))) = "3.56 h"`

(b) Calculate the cost

`"1 kWh" = "1kJ"/(1color(red)(cancel(color(black)("s")))) × 1 color(red)(cancel(color(black)("h"))) × (60 color(red)(cancel(color(black)("min"))))/(1 color(red)(cancel(color(black)("h")))) × (60 color(red)(cancel(color(black)("s"))))/(1 color(red)(cancel(color(black)("min")))) = "3600 kJ"`

∴ `"Cost" = "38 396" color(red)(cancel(color(black)("kJ")))× (1 color(red)(cancel(color(black)("kWh"))))/(3600 color(red)(cancel(color(black)("kJ")))) × "6.3 ¢"/(1 color(red)(cancel(color(black)("kWh")))) = "67 ¢"`