Question

100g water is kept at temperature ${{8}^{o}}C$ in a calorimeter whose water equivalent is 50g. if an ice cube of mass 2g at temperature ${{0}^{o}}C$ is dropped in the calorimeter assuming no heat is exchanged with the surroundings the final temperature of mixture will be (latent heat of fusion of ice = 80cal/g and specific heat capacity of water $1cal/{{g}^{o}}C$.
$\begin{aligned} & A{{.4.2}^{o}}C \\\ & B{{.6.8}^{o}}C \\\ & C{{.2.8}^{o}}C \\\ & D{{.0}^{o}}C \\\ \end{aligned}$

Answer 1

The heat gained by one system must be equal to the heat lost by the other. So, calculate heat gained and the heat lost by both the systems then evaluate final temperature by equating heat gain is equal to heat lost to get the required answer.

Complete answer:
Given data:
Water at ${{8}^{o}}C$ is 100g
The calorimeter water equivalent is 50g
So the total amount of water = $100+50=150g$
As we know, mass of ice = 2g at temperature ${{0}^{o}}C$
It is given that
The Latent heat of fusion of ice= $80cal/g$
The Specific heat capacity of water = $1cal/{{g}^{o}}C$
Now the heat required to melt the ice = $mL=2\times 80$ = 160cal
The heat required to melt the ice will be supplied by water. So, heat supplied by water = $mc\Delta t$
Where, m represents the total mass of the water
C represents the heat capacity of water
$\Delta t$ represents the change in temperature
Putting the given values in above formula, we get, the heat supplied by water = $\left( 150 \right)\left( 1 \right)\Delta T$
$\begin{aligned} & \Rightarrow 160=150\times 1\times \Delta T \\\ & \Rightarrow 160=150\times \Delta T \\\ & \Rightarrow \dfrac{160}{150}=\Delta T \\\ & \therefore \Delta T={{1.066}^{o}}C \\\ \end{aligned}$
Now, the final temperature of the water = $8-1.066={{6.93}^{o}}C$
Let the final equilibrium temperature be represented by T, therefore
Heat gained = heat lost
\eqalign{ & \left( 2 \right) \times \left( 1 \right) \times \left( {T - 0} \right) = \left( {150} \right) \times \left( 1 \right) \times \left( {T - 6.93} \right) \cr & \Rightarrow 2T = 150\left( {T - 6.93} \right) \cr & \Rightarrow T = 75\left( {T - 6.93} \right) \cr & \Rightarrow T = 75T - 75\left( {6.93} \right) \cr & \Rightarrow 75\left( {6.93} \right) = 74T \cr & \Rightarrow \dfrac{{517.5}}{{74}} = T \cr & \Rightarrow T = 6.99 \cr & \therefore T \approx 7^\circ C \cr}

So, the correct answer is “Option B”.

Note:
Specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of the unit mass of the given substance. Additionally, we need to remember the values of latent heat of fusion of ice and specific heat capacity of water.