Question #d5e47

Question

Question #d5e47

1 Answer

Impact of this question

3265 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-11T10:02:21

$3030 k J$ $3030kJ$

Explanation:

Your method is 100% correct, but you made a small calculation error in converting the lb to g, by a factor of 10, resulting in you final answer being out by a factor of 10.

The amount of energy required to melt $1 k g$ $1kg$ of ice at $0^{\circ} C$ $0^@C$ is equal to the specific latent heat of fusion for ice and has value $L_{f} = 334 k J / k g$ $L_f=334kJ//kg$ .

So for $20 \times 0.4536 = 9.072 k g$ $20xx0.4536=9.072kg$ , we get that energy required to melt it is
$W = m L_{f} = 9.072 k g \times 334000 J / k g$ $W=mL_f=9.072kgxx334000J//kg$

$= 3.03 M J$ $=3.03MJ$

$= 3030 k J$ $=3030kJ$

Science

Math

Humanities

... and beyond

Question #d5e47

1 Answer

Explanation:

Related questions

Impact of this question