Question

2kg of ice at $\text{- 2}{{\text{0}}^{\text{o}}}\text{C}$ is mixed with 5kg of water at $\text{2}{{\text{0}}^{\text{o}}}\text{C}$in an insulating vessel having negligible heat capacity. Calculate the final mass of water left in the container. Given: Specific of water and ice are $\text{1 kcal k}{{\text{g}}^{-1}}^{\text{o}}{{\text{C}}^{\text{-1}}}$ and $\text{0}\text{.5 kcal k}{{\text{g}}^{\text{-1}}}^{\text{o}}{{\text{C}}^{\text{-1}}}$ and latent heat of fusion of ice $\text{80 kcal k}{{\text{g}}^{-1}}^{\text{o}}{{\text{C}}^{\text{-1}}}$.
a) 7kg
b) 6kg
c) 4kg
d) 2kg

Answer 1

- Hint: It is given that the vessel in which the ice and water is contained has negligible heat capacity. Hence whatever the heat will be lost by water will change the properties of the ice. Hence using the principle of calorimetry we can determine how much of heat will be used to change ice into water and find the amount of water present in the container at equilibrium.

Complete step-by-step solution
To begin with the question let us first understand the principle of calorimetry. It states that heat lost by a body is equal to the heat gained by another body provided the heat is not lost into the surrounding.
The heat in the world always flows from higher temperature to lower temperature. Hence heat from water at 20 degree Celsius will be absorbed by ice at -20 degree Celsius.
Now we are ready to calculate the mass of the water remained in the container at thermal equilibrium
Heat lost by any body or heat gained by a body is given by $\Delta H=m{{c}_{P}}\Delta T...(1)$, where m is the mass of the body, ${{c}_{P}}$is the specific heat of the body and $\Delta T$is the change in temperature of the body. The heat required for ice to get converted into liquid from 0 degree Celsius if given by $\text{h = m}{{\text{L}}_{\text{FUS}}}...(2)$, where m is the mass of the ice and ${{\text{L}}_{\text{FUS}}}$ is the latent heat of fusion.
Let us first calculate the heat required by the ice to go from -20 to 0 degree Celsius. Using equation 1, we get,
$\Delta H=m{{c}_{ice}}\Delta T$
$\Delta H=2\times 0.5\times (20-0)=20\text{ kcal}$. This is the energy that is absorbed from the water hence the change in the temperature of the water is,
$\Delta H=m{{c}_{ice}}\Delta T$
$\begin{align}
  & \text{20=5 }\!\!\times\!\!\text{ 1 }\!\!\times\!\!\text{ (20 - T)} \\
 & \text{20 - T= 4, hence} \\
 & \text{T=1}{{\text{6}}^{\circ }}\text{C} \\
\end{align}$
Now let us say all the heat from the water till it reaches 0 degree Celsius is lost to the change the state of ice to water. Hence the lost of heat by water is equal to,
$\text{ }\!\!\Delta\!\!\text{ H = m}{{\text{c}}_{water}}\text{ }\!\!\Delta\!\!\text{ T}$
$\begin{align}
  & \text{ }\!\!\Delta\!\!\text{ H = 5}\times \text{1}\times (16-0) \\
 & \text{ }\!\!\Delta\!\!\text{ H=80 kcal} \\
\end{align}$
This heat will be used to convert ice to water at 0 degree Celsius. Using equation 2 let us calculate how much of ice will get converted to water i.e.
$\text{h = m}{{\text{L}}_{\text{FUS OF ICE}}}$
$\begin{align}
  & \text{80 kcal= m 80 kcal, hence} \\
 & m=1kg \\
\end{align}$
Now the container is in thermal equilibrium and hence that will be no exchange of heat between the ice and the water. Initially there was 5 kg of water in the container. After the exchange of heat 1 kg of ice got converted to water, hence the total mass of water in the container is equal to 6 kg.
Hence the correct answer is option B.

Note: All the heat is transferred to the ice so that no heat is lost into the surroundings. If further heat was still getting lost to the surrounding even after thermal equilibrium than water would have slowly turned into ice. This is because the temperature of the water would have decreased and would have been entirely converted to ice.