Question

Certain parts of the world don't have enough fresh drinking water. One possible solution is to tow icebergs from Antarctica and melt them as needed. If 6.10 ✕ 1011 kJ of energy were needed to melt an iceberg, how much did the iceberg weigh? The enthalpy of fusion (melting or freezing) for ice is 6.01 kJ/mol, and 1 metric ton = 1,000 kg. ____ metric tons

Answer 1

Answer:

Weight of the iceberg is $1.83\times 10^{6}$ metric ton

Explanation:

Molar mass of $H_{2}O$ = 18 g

So, 6.01 kJ of heat energy is needed to melt 18 g of ice

Hence, $6.10\times 10^{11}$ kJ of heat energy is needed to melt $\frac{18\times 6.10\times 10^{11}}{6.01}g$ of ice or $1.83\times 10^{12}$ g of ice

Now, $10^{3}kg$ = 1 metric ton

So, $10^{6}$ g = 1 metric ton   ( 1 kg = 1000 g)

Hence, $1.83\times 10^{12}$ g =$\frac{1\times 1.83\times 10^{12}}{10^{6}}$ metric ton = $1.83\times 10^{6}$ metric ton

So, weight of the iceberg is $1.83\times 10^{6}$ metric ton