Question

25.5 g of liquid Benzene (`C_6H_6`) loses 200 J of heat as it cools. What is the temperature decrease?

Cp(`C_6H_6`) = 1740 J/`kg*C`

Answer 1

The temperature decrease is `~~"5"^@"C"`.

Explanation:

Use the equation:

`q=mcDeltaT`,

where:

`q` is energy, `m` is mass, `c` is specific heat capacity, and `DeltaT` is the change in temperature.

Known

`q="200 J"`

`m="25.5 g"="0.0255 kg"`
(Mass needs to be in kg due to the units used for the specific heat capacity.)

`c_"C6H6"=(1740"J")/("kg"*""^@"C")`

Unknown

`DeltaT`

Solution

Rearrange the equation to isolate `DeltaT`. Plug in the known values and solve.

`DeltaT=q/(m*c)`

`DeltaT=(200color(red)cancel(color(black)("J")))/((0.0255color(red)cancel(color(black)("kg")))xx(1740color(red)cancel(color(black)("J")))/(color(red)cancel(color(black)("kg"))*""^@"C"))="5"^@"C"` (rounded to one significant figure)

The temperature decrease is `~~"5"^@"C"`.