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

Weight of the iceberg is
`1.83* 10^(6)` metric ton

Step-by-step 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* 10^(11)` kJ of heat energy is needed to melt
`(18* 6.10* 10^(11))/(6.01)g` of ice or
`1.83* 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* 10^(12)` g =
`(1* 1.83* 10^(12))/(10^(6))` metric ton =
`1.83* 10^(6)` metric ton

So, weight of the iceberg is
`1.83* 10^(6)` metric ton