Calorimetry
4.0 Latent heat of fusion and vaporisation
4.0 Latent heat of fusion and vaporisation
Apart from raising the temperature of the body, heat supplied to a body may cause a change in phase. Such as solid to liquid or liquid to vapour.
During this, the temperature remains constant.
The amount of heat needed to melt a substance of mass $m$ at constant temperature can be written as $$Q = mL$$ where $L$ is the latent heat of fusion.
It can also be defined as the amount of heat required to change the unit mass of a solid into liquid at constant temperature. $$L = \frac{Q}{m}$$ Latent heat of fusion of ice =$80\,cal\,{g^{ - 1}}$
Latent heat of vaporisation
The amount of heat required to change the unit mass of a liquid into its gaseous state at constant temperature is called latent heat of vaporisation.$$Q = mL$$Latent heat of vaporisation of water is $540\,cal\,{g^{ - 1}}$.
Question 3. A piece of ice of mass 100$g$ and at temperature ${0^o}C$ is put in 200$g$ of water at ${25^o}C$. Hpw much ice will melt as the temperature of the water reaches ${0^o}C$? The specific heat of water = $4200\,J\,k{g^{ - 1}}\,{K^{ - 1}}$ and the specific latent heat of fusion = $340000\,J\,k{g^{ - 1}}\,{K^{ - 1}}$.
Solution:
The heat released as the water cools down from ${25^o}C$ to ${0^o}C$ is $$\begin{equation} \begin{aligned} Q = ms\Delta T = \left( {0.2kg} \right)\left( {4200\,J\,k{g^{ - 1}}\,{K^{ - 1}}} \right)\left( {25\,K} \right) \\ Q = 21000\,J \\\end{aligned} \end{equation} $$The amount of ice melted by this much of heat is given by$$m = \frac{Q}{L} = \frac{{21000\,J}}{{3.4 \times {{10}^5}\,J\,k{g^{ - 1}}}} = 62\,g$$.
Question 4. The temperature of equal masses of three of three different liquids A, B, C are ${12^o}C$, ${19^o}C$ and ${28^o}C$. The temperature when A and B are mixed is ${16^o}C$, when B and c are mixed it is ${23^o}C$. What should be the temperature when A and C are mixed?
Solution:
Let $m$ be the mass of each liquid and ${s_A}$, ${s_B}$ and ${s_C}$ be the specific heats of the liquids.
When A and B are mixed, the final temperature is ${16^o}C$
Heat gained by A = Heat lost by B$$\begin{equation} \begin{aligned} m{s_A}\left( {16 - 12} \right) = m{s_B}\left( {19 - 16} \right) \\ {s_B} = \frac{4}{3}{s_A} \\\end{aligned} \end{equation} $$When B and C are mixed.Heat gained by B = Heat lost by C$$\begin{equation} \begin{aligned} m{s_B}\left( {23 - 19} \right) = m{s_C}\left( {28 - 23} \right) \\ {s_C} = \frac{4}{5}{s_B} \\\end{aligned} \end{equation} $$From the above two equations$${s_C} = \frac{4}{5} \times \frac{4}{3}{s_A} = \frac{{16}}{{15}}{s_A}$$When A and C are mixed, let the final temperature be $T$,$$\begin{equation} \begin{aligned} m{s_A}\left( {T - 12} \right) = m{s_C}\left( {28 - T} \right) \\ T - 12 = \frac{{16}}{{15}}\left( {28 - T} \right) \\ T = 20.26{}^oC \\\end{aligned} \end{equation} $$
