# In a calorimeter, 1.0 kg of ice melts at 0°C. The enthalpy of fusion of the ice is 334 J/g. How much heat was absorbed?

##### 1 Answer

#### Answer:

#### Explanation:

A given substance's *enthalpy of fusion*, **one gram** of that substance from *solid* at its melting point to *liquid* at its melting point.

It's important to remember that **phase changes** take place at **constant temperature**, which is why you'll sometimes see the enthalpy of fusion being referred to as the *latent heat of fusion*.

So, ice has an enthalpy of fusion equal to

To determine how much heat is needed to convert *conversion factor*.

First, convert the mass of the sample to *grams* by using the conversion factor

#"1 kg" = 10^3"g"#

This will get you

#1.0 color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 1.0 * 10^3"g"#

Use the mass of the sample and the enthalpy of fusion of ice to find

#1.0 * 10^3color(red)(cancel(color(black)("g"))) * overbrace("334 J"/(1color(red)(cancel(color(black)("g")))))^(color(purple)(DeltaH_"fus")) = 334 * 10^3"J"#

You need to round this off to two **sig figs**, the number of sig figs you have for the sample of ice. You can also express the result in *kilojoules* by using the conversion factor

#"1 kJ" = 10^3"J"#

The answer will thus be

#"heat absorbed" = color(green)(|bar(ul(color(white)(a/a)"330 kJ"color(white)(a/a)))|)#