# Question #23181

Nov 15, 2016

$\text{70.1 g}$ of water.

#### Explanation:

Ok, so I did a quick search and found the heat capacity for Cu.

The formula for thermal energy exchange is:

${Q}_{\text{released" + Q_"absorbed}} = 0$

${\left(m c \Delta T\right)}_{\text{released" + (mcDeltaT)_ "absorbed}} = 0$

Where

• $Q$ is the quantity of heat
• $m$ is the mass in kg,
• $c$ is the specific heat capacity
• $\Delta T$ is the change in temperature, $\left({T}_{2} - {T}_{1}\right)$.

Released is the Copper and the absorbed is the water.

${\left(m c \Delta T\right)}_{\text{Cu" + (mcDeltaT)_ "water}} = 0$

$0.02 \left(380\right) \left(22 - 100\right) + m \left(4180\right) \left(22 - 20\right) = 0$

$- 592.8 + 8360 m = 0$

$8360 m = 592.8 \implies m = 0.071$

Since

$\text{0.071 kg " = " 70.1 g}$

you have $\text{70.1 g}$ of water.

Hope this helps :)