Illustration :

The ratio of radii of two copper spheres is \(2: 1\) and they are kept at same temperature. The ratio of their heat capacities will be -

(1) \(2: 1\)

(2) \(1: 1\)

(3) \(8: 1\)

(4) \(4: 1\)

Sol. \(\frac{\mathrm{dQ}_{1}}{\mathrm{dQ}_{2}}=\frac{\frac{4}{3} \pi \mathrm{r}_{1}^{3} \rho \mathrm{S}}{\frac{4}{3} \pi \mathrm{r}_{2}^{3} \rho \mathrm{S}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3}=\left(\frac{2}{1}\right)^{3}=\frac{8}{1}\)

Illustration :

\(5 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\) is dropped ina beaker containing \(20 \mathrm{~g}\) of water at \(40^{\circ} \mathrm{C}\), then

(1) All the ice will not melt into water

(2) All the ice will melt and the resulting temperature of water will be \(0^{\circ} \mathrm{C}\)

(3) All the ice will melt and the resulting temperature of water will be \(25^{\circ} \mathrm{C}\)

(4) All the ice will melt and the resulting temperature of water will be \(16^{\circ} \mathrm{C}\)

Sol. Heat required to melt \(5 \mathrm{~g}\) ice \(=5 \times 80=400 \mathrm{cal}\). Heat available with \(20 \mathrm{~g}\) water at \(40^{\circ} \mathrm{C}=20 \times 1 \times 40=800\) cal out of this, 400 cal heat will be used to melt the ice and remaining heat will raise the temperature. Thus Heat given by water \(=\) Heat taken by ice 

\( 20 \times 1 \times(40-\theta)=5 \times 80+5 \times 1 \times(\theta-0) \)

 \(800-20 \theta=400+5 \theta \)

\( \Rightarrow \quad 25 \theta=400 \)

\(\text { or } \quad \theta=\frac{400}{25}=16^{\circ} \mathrm{C}\)

Illustration :

A \(5 \mathrm{~g}\) piece of ice at \(-20^{\circ} \mathrm{C}\) is put into \(10 \mathrm{~g}\) of water at \(30^{\circ} \mathrm{C}\). Assuming that heat is exchanged only between the ice and the water, find the final temperature of the mixture. Specific heat capacity of ice \(=2100 \mathrm{~J} / \mathrm{kg}^{\circ}{ }^{\circ} \mathrm{C} \mathrm{specific}\) heat capacity of water \(=4200 \mathrm{~J} / \mathrm{kg}-{ }^{\circ} \mathrm{C}\) and latent heat of fusion of ice \(=3.36 \times 10^{5} \mathrm{~J} / \mathrm{kg}\).

Sol. The heat given by the water when it cools down from \(30^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\) is

\((0.01 \mathrm{~kg})\left(4200 \mathrm{~J} / \mathrm{k}-{ }^{\circ} \mathrm{C}\right)\left(30^{\circ} \mathrm{C}\right)=1260 \mathrm{~J}\)

The heat required to bring the ice to \(0^{\circ} \mathrm{C}\) is \((0.005 \mathrm{~kg})\left(2100 \mathrm{~J} / \mathrm{kg}-{ }^{\circ} \mathrm{C}\right)\left(20^{\circ} \mathrm{C}\right)=210 \mathrm{~J}\). The heat required to melt \(5 \mathrm{~g}\) of ice is \((0.005 \mathrm{~kg})\left(3.36 \times 10^{5} \mathrm{~J} / \mathrm{kg}\right)=1680\).

We see that whole of the ice cannot be melted as the required amount of heat is not proved by the water. Also, the heat is enough to bring the ice to \(0^{\circ} \mathrm{C}\). Thus the final temperature of the mixture is \(0^{\circ} \mathrm{C}\) with some of the ice melted.

Illustration :

\(1 \mathrm{~g}\) of steam at \(100^{\circ} \mathrm{C}\) can melt how much ice at \(0^{\circ} \mathrm{C}\) ? Latent heat of ice \(=80 \mathrm{cal} / \mathrm{g}\) and latent heat of steam \(=540 \mathrm{cal} / \mathrm{g}\).

Sol. Heat required by ice for melting of \(m \mathrm{~g}\) of ice \(=m L=m \times 80 \mathrm{cal}\)

Heat available with steam for being condensed and then brought to \(0^{\circ} \mathrm{C}\)

\( =1 \times 540 \times 100 \)

\(=640 \mathrm{cal} \)

\(m  \times 80=640\)

\(  \text { or } \quad m=\frac{640}{80}=8 \text { grams } \)

Illustration :

A tap supplies water at \(10^{\circ} \mathrm{C}\) and another tap at \(100^{\circ} \mathrm{C}\). How much hot water must be taken so that we get \(20 \mathrm{~kg}\) of water at \(35^{\circ} \mathrm{C}\) ?

Sol. Let mass of hot water \(=m \mathrm{~kg}\)

\( \text { mass of cold water } \)

 \(=(20-m) k g \)

 \(\text { Heat taken by cold water } \)

\( =(20-m) \times 1 \times(35-10) \)

 \(\text { Heat given by hot water } \)

\( =m \times 1 \times(100-35)\)

Law of mixture gives

Heat given by hot water

\( =\text { Heat taken by cold water } \)

\( m \times 1 \times(100-35)=(20-m) \times 1 \times(35-10) \)

 \(65 m=(20-m) \times 25 \)

 \(65 \mathrm{~m}=500-25 \mathrm{~m}\)

or

 \(90 m=500 \)

\(m=\frac{500}{90}=5.56 \mathrm{~kg}\)

Illustration :

\(5 \mathrm{~g}\) ice of \(0^{\circ} \mathrm{C}\) is mixed with \(5 \mathrm{~g}\) of steam at \(100^{\circ} \mathrm{C}\). What is the final temperature?

Sol. Heat required by ice to raise its temperature to \(100^{\circ} \mathrm{C}\),

\( Q_{1}=m_{1} L_{1}+m_{1} c_{1} \Delta \theta_{1} \)

\( =5 \times 80+5 \times 1 \times 100 \)

 \(=400+500=900 \mathrm{cal}\)

Heat given by steam when condensed,

\( \mathrm{Q}_{2}=\mathrm{m}_{2} \mathrm{~L}_{2} \)

\(=5 \times 536=2680 \mathrm{cal}\)

As \(\mathrm{Q}_{2}>\mathrm{Q}_{1}\). This means that whole steam is not even condensed.

Hense temperature of mixture will remain at \(100^{\circ} \mathrm{C}\)