Question

Two 50g ice cubes are dropped into 250g of water into a glass. If the water was initially at a temperature of $25{}^\circ C$ and the temperature of the ice $-15{}^\circ C$. Find the final temperature of water. Final amount of water and ice. $(\text{Specific heat of ice=0}\text{.5}cal/g{}^\circ C\text{ and }L\text{=80}cal/g\text{)}$

Answer 1

Here two ice cubes are dropped into a glass of water and initial temperature of ice and water are given. As ice cubes are dropped after a while they will start melting and there will be a change in the temperature of water and we have to find the final temperature of water. We have also given the specific heat and latent heat from which we can calculate the heat absorbed by the ice.

Formula used:
$Q=ms\Delta T$

Complete answer:
When ice cubes are dropped into water, ice will melt therefore there will be heat loss by water and heat will be gained by the ice cubes.
Let us consider some amount of ice, say mass m, converts into water. It will have latent heat while changing the state from solid to liquid and also there will be amount of heat gain while the temperature changes from $-15{}^\circ C$ to $0{}^\circ C$ of ice, after $0{}^\circ C$ ice starts melting again there will be gain of heat while temperature changes from $0{}^\circ C$ to final temperature ${{T}_{f}}$ for the amount of ice which has been converted into water.
Hence the total amount of heat gained by the ice will be given as
${{Q}_{G}}={{m}_{i}}{{s}_{i}}\Delta {{T}_{i}}+mL+m{{s}_{w}}\Delta T'$
Where ${{m}_{i}}$ is the mass of ice cube, ${{s}_{i}}$ is the specific heat of ice, $\Delta {{T}_{i}}$is change in temperature of ice, L is latent heat, ${{s}_{w}}$ is the specific heat of water and $\Delta T'$ is the change in temperature of ice which has been converted into water.
Given ${{m}_{i}}=2\times 50g=100g,{{s}_{i}}=0.5cal/g{}^\circ C,\Delta {{T}_{i}}=0-(-15)=15{}^\circ C, L=80cal/g, {{s}_{w}}=1cal/g, \Delta T'={{T}_{f}}-0$
Substituting values in above formula we get

& {{Q}_{G}}=\left( 100 \right)\left( 0.5 \right)\left( 0-(-15) \right)+m\left( 80 \right)+m\left( 1 \right)\left( {{T}_{f}}-0 \right) \\\ & {{Q}_{G}}=50\left( 15 \right)+80m+m{{T}_{f}} \\\ & {{Q}_{G}}=750+80m+m{{T}_{f}} \\\ \end{aligned}$$ Now as ice is dropped into water, water will loss heat while ice is melting and heat will be released, then the amount of heat lost by water will be given as $${{Q}_{L}}={{m}_{w}}{{s}_{w}}\Delta {{T}_{w}}$$ Where $${{m}_{w}}$$is the mass of water, $${{s}_{w}}$$is the specific heat of water and $$\Delta {{T}_{w}}$$ Given $${{m}_{w}}=250g\text{ and }\Delta {{T}_{w}}=25{}^\circ C-{{T}_{f}}$$ Substituting values in above equation, we get (initial temperature of water is$$25{}^\circ C$$) $$\begin{aligned} & {{Q}_{L}}=\left( 250 \right)\left( 1 \right)\left( 25-{{T}_{f}} \right) \\\ & {{Q}_{L}}=6250-250{{T}_{f}} \\\ \end{aligned}$$ According to law of conservation of heat, heat loss will be equal to heat gain $$\begin{aligned} & {{Q}_{L}}={{Q}_{G}} \\\ & 6250-250{{T}_{f}}=750+80m+m{{T}_{f}} \\\ & m{{T}_{f}}+250{{T}_{f}}=6750-750-80m \\\ & {{T}_{f}}(m+250)=5500-80m \\\ \end{aligned}$$ If one ice cubes was converted then, $$m=50g$$ $$\begin{aligned} & {{T}_{f}}(50+250)=5500-80\left( 50 \right) \\\ & 300{{T}_{f}}=5500-4000 \\\ & {{T}_{f}}=\dfrac{1500}{300} \\\ & {{T}_{f}}=5{}^\circ C \\\ \end{aligned}$$ The final temperature is $$5{}^\circ C$$and the final amount of water and ice will be 300g and 50g as one ice cube is converted into water which is 50g therefore 50g will be added in the mass of water. **Note:** Here we have given the latent heat of fusion as ice will be changing into water i.e. from solid to liquid whereas when liquid changes into solid latent heat is called as latent heat of evaporation. Latent heat basically tells us the amount of heat required absorbed or released per unit mass while changing the state.