Question

Ice at $- {20^ \circ }C$ is added to $50{\text{ }}g$ of water at ${40^ \circ }C$. When the temperature of the mixture reaches ${0^ \circ }C$, it is found that $20{\text{ }}g$ of ice is still unmelted. The amount of ice added to the water was close to
(Specific heat of water$= 4.2\dfrac{J}{{{g^ \circ }C}}$, Specific heat of ice$= 2.1\dfrac{J}{{{g^ \circ }C}}$ and Heat of fusion of water at ${0^ \circ }C$ $= 334{\text{ }}\dfrac{J}{g}$)
A. $50{\text{ }}g$
B. $100{\text{ }}g$
C. $60{\text{ }}g$
D. $40{\text{ }}g$

Answer 1

By using the Law of Calorimetry, we will find the amount of heat that is lost by the water by using the specific heat being given. Clearly, the amount of heat that is lost by water is gained by ice to lower its temperature to zero and also convert a part of it into water. By the same way, we will find the heat gained by ice and thus comparing both equations we will find the answer.

Complete step by step answer:
The amount of heat that is being lost by water, when water-ice mixture reaches ${0^ \circ }C$ is given as,
$Q = mc\Delta T - - - - \left( 1 \right)$
The variables are defined as-
$Q =$ amount of heat lost or gained.
$m =$ mass of the substance.
$c =$ specific heat of the substance.
$\Delta T =$ change in temperature.
The quantities that are already given in the question are-
$m = 50$, $c = 4.2$ and $\Delta T = 40 - 0 = 40$.
Substituting the values in equation $\left( 1 \right)$ we get the amount of heat lost by water is,
$Q = 50 \times 40 \times 4.2 = 8400{\text{ }}J$

Let the mass of ice be $m'$.Now using the equation $\left( 1 \right)$ , the amount of heat gained for the ice to convert from $- {20^ \circ }C$ to ${0^ \circ }C$ is
$Q = m' \times 2.1 \times 20 = 42m'$
According to the question only $20{\text{ }}g$ of ice is unmelted and the rest mass of ice is being melted by the heat into water. So, the mass that is melted is $m' - 20$ and the heat required be $Q'$. Amount of heat required to convert it into water is given by the formula of Latent heat of Fusion,
$Q = m{L_f} - - - - - - \left( 2 \right)$
The variables defined here are,
$Q =$ amount of heat lost or gained.
$m =$ mass of the substance.
${L_f} =$ Latent heat of fusion.

Given ${L_f} = 334{\text{ }}\dfrac{J}{g}$
Substituting the values in equation $\left( 2 \right)$ we get,
$Q' = \left( {m' - 20} \right) \times 334$
Therefore, we conclude that the heat required for the ice to lower its temperature to ${0^ \circ }C$ and to convert some part of it into water is provided by the heat lost by the water for its conversion from ${40^ \circ }C$ to ${0^ \circ }C$.
Thus, $8400 = \left( {m' - 20} \right) \times 334 + 42m'$
Simplifying the equation we get,
$376m' = 15080$
$\therefore m' = 40.10$
So, the amount of ice that was given to water is close to $40{\text{ }}g$.

Hence, the correct option is D.

Note: The heat energy that is being lost by the water due to its change in temperature while in the ice-water mixture, is equal to the heat energy that is being gained by the ice. It is because of the fact that energy can neither be created nor destroyed, it just transforms. Latent heat of a substance is the amount of heat that is required to change the state of a substance at approximately the same temperature for example from solid to liquid or liquid to gas. Specific heat of the material of a substance is defined as the amount of heat required to change the temperature of unit mass of a substance by ${1^ \circ }C$.