Question

The enthalpy of fusion for ice is $334{\text{ J/g}}$. How much heat is absorbed if $10.0$ grams of ice melts at ${0^\circ }{\text{C}}$ ?
A.$3340{\text{ J}}$
B.$334{\text{ J}}$
C.$33400{\text{ J}}$
D.$10{\text{ J}}$

Answer 1

To solve this question, we will use the formula of heat of fusion that relates heat energy to heat of fusion and mass. It is given as-
$\Rightarrow {\text{q = m}}{{.}}\Delta {{\text{H}}_{\text{f}}}$
Where q is the heat of fusion, m is the mass of the matter, and ${{\Delta }}{{\text{H}}_{\text{f}}}$ is the enthalpy of fusion. The heat of fusion is the amount of heat energy required when the state of matter is changed from solid to liquid. Put the given values in the formula and solve it to get the answer. Here we have to remember that the units should be the same to get the correct answer.

Complete step-by-step answer: Given, the enthalpy of fusion for ice $\Delta {{\text{H}}_{\text{f}}}{\text{ = 334 J/g}}$
The mass of ice that melts at ${0^\circ }{\text{C}}$(m) = $10.0$ grams
We have to find the absorbed heat.
Here since ice melts into the water so the phase of matter changes from solid to liquid so we will use the formula of the heat of fusion which is given as-
$\Rightarrow {\text{q = m}}{{.}}\Delta {{\text{H}}_{\text{f}}}$
Where q is heat of fusion, m is the mass of the matter and ${{\Delta }}{{\text{H}}_{\text{f}}}$ is the enthalpy of fusion
On putting the given value in the formula, we get-
$\Rightarrow {\text{q = 10 g}} \times 334{\text{ J/g}}$
On multiplication, we get-
$\Rightarrow {\text{q = }}334{\text{0 J}}$ {Here the unit gram gets cancelled}
Hence the heat absorbed is $3340{\text{ J}}$.

The correct answer is option A.

Note: Here temperature is not used in the formula because the temperature does not change when the state of matter is changed. Also, in this type of question, remember that heat of fusion is always a positive value (Helium is the exception as it gives a negative value) so if the answer is a negative value, recheck the solution.