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

## Cp(${C}_{6} {H}_{6}$) = 1740 J/$k g \cdot C$

Jun 10, 2018

The temperature decrease is $\approx \text{5"^@"C}$.

#### Explanation:

Use the equation:

$q = m c \Delta T$,

where:

$q$ is energy, $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the change in temperature.

Known

$q = \text{200 J}$

$m = \text{25.5 g"="0.0255 kg}$
(Mass needs to be in kg due to the units used for the specific heat capacity.)

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

Unknown

$\Delta T$

Solution

Rearrange the equation to isolate $\Delta T$. Plug in the known values and solve.

$\Delta T = \frac{q}{m \cdot c}$

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 $\approx \text{5"^@"C}$.