Question

Question #98fe4

Answer 1

You will need to add 9.4 kJ of heat.

Explanation:

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

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

where

`mcolor(white)(ll)` is the mass
`c color(white)(ml)` is the specific heat capacity
`ΔT` is the change in temperature

In your problem,

`m color(white)(ll)= "27 g"`
`c color(white)(ml)= "4.184 J·°C"^"-1""g"^"-1"`
`ΔT = "83 °C"`

∴ `q = 27 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1"·"g"^"-1"))) × 83 color(red)(cancel(color(black)("°C"))) = "9400 J" = "9.4 kJ"`