Question

The latent heat of fusion water is $334J/g$. How many grams of ice at ${0^ \circ }C$will melt by the addition of $3.34kJ$ of heat energy?

Answer 1

Hint: Latent heat can be defined as the energy or heat that is released or absorbed during a phase change of the substance and it is also related to a heat property known as enthalpy. By using the latent heat of fusion theory, the given question can be solved.
Formula used:
$ \Rightarrow Q = m{L_f}$
Where,
$Q$ is the heat supply,
$m$ is the mass of the substance,
${L_f}$ is the latent heat of the fusion.

Complete step by step answer:
The heat that is consumed or discharged when the matter melts or when it changes the state from strong matter to the fluid-structure under the constant temperature is known as the latent heat of fusion.
The latent heat has the enthalpy of the fusion fact during the softening the vitality of the heat is expected to change the nature of the substance under the pressure of the air from the solid to liquid. The temperature will remain stable or steady throughout the procedure.
The given problem can be solved with the help of the formula that is used to find the value of the heat supplied $Q$. It can be found by multiplying the value of the mass of the substance and the latent heat of the fusion. It can be represented by,
$ \Rightarrow Q = m{L_f}$
Where,
$Q$ is the heat supply,
$m$ is the mass of the substance,
${L_f}$ is the latent heat of the fusion.
The mass of the substance is the value that is needed to be found. The equation can be rewritten as,\[ \Rightarrow m = \left( {\dfrac{Q}{{{L_f}}}} \right)\]
Let us try to solve the given problem. The value of the heat supplied is given. That is the value of $Q$is given as $3.34kJ$. The value of the latent heat of fusion of water is $334J/g$.
We have the formula,
\[ \Rightarrow m = \left( {\dfrac{Q}{{{L_f}}}} \right)\]
Substitute the values in the formula.
\[ \Rightarrow m = \left( {\dfrac{{3.34kJ}}{{334J/g}}} \right)\]
First, convert the kilojoules into joules. That is,
\[ \Rightarrow m = \left( {\dfrac{{3.34 \times {{10}^3}J}}{{334J/g}}} \right)\]
Cancel out the common terms we get,
\[ \Rightarrow m = \left( {\dfrac{{3.34 \times {{10}^3}}}{{334g}}} \right)\]
Using division for simplification. We get,
\[ \Rightarrow m = \left( {\dfrac{{3.34 \times {{10}^3}}}{{334g}}} \right)\]
\[ \Rightarrow m = 10g\]
Therefore, $10g$ of ice at ${0^ \circ }C$ will melt by the addition of $3.34kJ$of heat energy.

Note:
An important point to remember is that when it comes to latent heat the temperature will always remain constant. If the mechanism is concerned, the latent heat work is needed to overcome the attractive force that is used to hold the molecules and the atoms together as a substance.