Question

Calculate the mass of ice required to lower the temperature of $300\,{\text{g}}$ water at $40\,^\circ {\text{C}}$ to water at $0\,^\circ {\text{C}}$. (Specific latent heat of ice $= 336\,{\text{J/g}}$, specific heat capacity of water $= 4.2\,{\text{J/g}}^\circ {\text{C}}$ ).

Answer 1

First of all, we will find the heat absorbed by the ice and heat lost by the water. Use the idea that the heat lost by the water is absorbed by ice. Manipulate the equation and simplify accordingly to obtain the result.

Complete step by step answer:
In the given problem, mass of water is $300\,{\text{g}}$ .
Initial temperature of water is $40\,^\circ {\text{C}}$ .
Final temperature of water is $0\,^\circ {\text{C}}$ .
Specific latent heat of ice $= 336\,{\text{J/g}}$ .
Specific heat capacity of water $= 4.2\,{\text{J/g}}^\circ {\text{C}}$ .
Before we begin, we will highlight the fact that ice is added to the water to lower its temperature. So, the question arises that where the heat will go, as we know, that heat is also a form of energy. According to the law of conservation of energy, energy can neither be destroyed nor be created, rather it can be converted to some other forms. Here, in this case the heat lost by the water to lower its temperature is actually used to melt the ice which is added. This is the roadmap for the problem.
Heat energy absorbed by the ice is given by the formula:
$H = ML$ …… (1)
Where,
$H$ indicates heat absorbed.
$M$ indicates mass of ice required.
$L$ indicates specific latent heat of ice.
Heat released by the water is given by the formula:
$H = mc\theta$ …… (2)
Where,
$H$ indicates heat released.
$m$ indicates mass of water.
$c$ indicates specific heat capacity of water.
$\theta$ indicates drop in temperature.
Comparing equations (1) and (2), we get:
$ML = mc\theta$ …… (3)
Substituting the required values in the equation (3), we get:
$M \times 336 = 300 \times 4.2 \times \left( {40 - 0} \right) \\\ \Rightarrow M = \dfrac{{300 \times 4.2 \times 40}}{{336}} \\\ \therefore M = 150\,{\text{g}} \\\$
Hence, the required mass of ice is $150\,{\text{g}}$.

Note: While solving this problem, always keep in mind that, magnitude of both the heat energies are equal, as the heat released by the water is used to melt the ice and lower the temperature. It is important to note that $\theta$ is the drop/fall/range of temperature, where the students tend to mess up this with a single reading of temperature. You can take the mass in kilograms also, that is not an issue, as the answer will remain the same.