Question

A sample of mercury absorbed 257 J of heat and its mass was .45 kg. If it's temperature increased by 4.09 K, what is its specific heat in J° / kg K?

Answer 1

The specific heat formula is:

`c = Q/(m × ΔT)`

Where:

`c`: specific heat, in J/(kg.K)

`Q`: heat required for the temperature change, in J

`ΔT`: temperature change, in K

`m`: mass of the object, in kg

For our equation, we are given everything in the right units, so all we need is to plug in the numbers.

`c = Q/(m × ΔT)`

`c = (257)/(.45 xx 4.09)`

`c = 139.64 (J)/(kgxxK)`

Answer 2

The specific heat of mercury to two significant figures is `"140 J"/("kg"*"K")`.

Explanation:

Use the equation `color(red)(Q=mcDeltat)`, where `Q` is energy gained or lost in Joules; `m` is mass, in this case kg; `c` is specific heat, in this case in `"J"/("kg"*"K")"`; and `Deltat` is change in temperature, in this case Kelvins.
Determine what variables are known, and unknown. Then solve the equation for the unknown variable.

Known
`Q="257 K"`
`m="0.45 kg"`
`Deltat="4.09 K"`

Unknown
`c_"Hg"`

Solution
Rearrange the equation to isolate `c_"Hg"`. Substitute the known values into the equation and solve.

`c_"Hg"=(Q)/(m*Deltat)=(257"J")/((0.45"kg")xx(4.09"K"))="140 J"/("kg"*"K")` rounded to two signficant figures