4. A mixture of 500 g water and 100 g ice at 0^oC is kept in a copper calorimeter of mass 200 g. How much steam from the boiler be passed to the mixture so that the temperature of the mixture reaches 40^oC? (Latent heat of fusion of ice = 3.36 x 10⁵ JKg^-1, specific heat capacity of copper = 400 Jkg^{-1 o}C^-1, latent heat of vaporization of steam = 2.26 x 10⁶ Jkg^-1).

Solution
Given,
Mass of water (m_w) = 500 g = 0.5 kg
Mass of ice (m_i) = 100 g = 0.1 kg
Temperature [latex](\theta_1[/latex]) = 0^oC
Mass of calorimeter (m_c) = 200 g = 0.2 kg
Temperature of mixture [latex](\theta[/latex]) = 40^oC
Mass of steam (m_s) =?
We have, heat gained by ice, water and calorimeter
Q_gained = (m_iL_f + m_is_w [latex]\Delta \theta[/latex]) + (m_ws_w [latex]\Delta \theta[/latex] + (m_cs_c [latex]\Delta \theta[/latex])
= (0.1 x 3.36 x 10⁵ + 0.1 x 4200 x 40) + (0.5 x 4200 x 40) + (0.2 x 400 x 40)
= 1.376 x 10⁵ J
Heat lost by steam
Q_lost = (m_sL_v + m_ss_w )
= m_s x 2.26 x 10⁶ + m_s x 4200 x (100 – 40)
= (2.26 x 10⁶ + 2.52 x 10⁵)m_s
= 2.512 x 10⁶m_s
According to principle of calorimetry,
Q_gained = Q_lost
Or, 1.376 x 10⁵ = 2.512 x 10⁶m_s
Or, m_s = [latex]\frac{1.376\times 10^5}{2.512\times 10^6}[/latex]
= 0.05477 kg            
= 54.77 gm.

16. What is the result of mixing 100 g of ice at 0^oC and 100 g of water at 100^oC. Latent heat of fusion of ice = 336 x 10³ JK^-1, specific heat of water = 4200 Jkg^-1K^-1.

Solution
Given,
Mass of ice (m_i) = 100 gm
Mass of water (m_w) = 100 gm

Amount of heat required to melt ice (Q₁) = m_iL_f
            = 100 x 80
                = 8000 Cal
Now, amount of heat lost by water and vessel i.e. amount of heat available for melting ice (Q₂) = m_ws_w
                = 100 x 1 x 100
= 10000 Cal
Since, Q₂ > Q₁, all ice melts, and the resulting temperature rises to [latex]\theta^oC[/latex]
Again, let’s have a look on the following figure.

Heat gained by ice on changing from 0^oC to [latex]\theta^oC[/latex],
Q_gained = m_iL_f + m_is_w
= 100 x 80 + 100 x 1 x [latex]\theta[/latex]
= 8000 + 100[latex]\theta[/latex]
Heat lost by water from 100^oC to [latex]\theta^oC[/latex],
Q_lost = m_ws_w
= 100 x 1 x (100 -[latex]\theta[/latex])
= 10000 – 100[latex]\theta[/latex]
From the principle of calorimetry,
Heat lost = Heat gained
i.e. 10000 – 100  = 8000 + 100
Or, 10000 – 8000 = 100
Or, 2000 = 200[latex]\theta[/latex]
Or, [latex]\theta = \frac{2000}{200} = 10^oC[/latex]
So, in the mixture,
Mass of water formed = 100 + 100 = 200 gm
Mass of ice remaining = 100 – 100 = 0 gm
And, resulting temperature of mixture = 10^oC.