# Question #38eda

Feb 21, 2017

$\textsf{7.9}$

#### Explanation:

Use the idea that the heat given out by the aluminium = the heat taken in by the water.

I notice that the heat capacities are given per mole so we need to work out the number of moles of aluminium.

I will use $\textsf{{A}_{r} \left[A l\right] = 27}$

$\therefore$$\textsf{{n}_{A l} = \frac{m}{A} _ r = \frac{60}{27} = 2.22 \textcolor{w h i t e}{x} \text{mol}}$

The heat given out by the aluminium =$\textsf{{n}_{A l} \times {C}_{A l} \times \Delta T}$

$\textsf{\Delta T = 75 - 20 = 55 \textcolor{w h i t e}{x} \text{deg}}$

$\therefore$Heat given out = $\textsf{2.22 \times 24.3 \times 55 = 2967 \textcolor{w h i t e}{x} J}$

This heat is taken in by the water.

$\therefore$$\textsf{2967 = {n}_{{H}_{2} O} \times {C}_{{H}_{2} O} \times \Delta T}$

In this case $\textsf{\Delta T = 20 - 15 = 5 \textcolor{w h i t e}{x} \text{deg}}$

$\therefore$$\textsf{2967 = {n}_{{H}_{2} O} \times 75.3 \times 5}$

$\therefore$$\textsf{{n}_{{H}_{2} O} = \frac{2967}{75.3 \times 5} = 7.88 \textcolor{w h i t e}{x} \text{mol}}$