91Ó°ÊÓ

1. Mass of tea, ${\mathrm{M}}_{\mathrm{W}}=500\mathrm{g}$
2. Specific heat of water ${\mathrm{C}}_{\mathrm{W}}=4190\mathrm{J}/\mathrm{kg}.{}^{\mathrm{o}}\mathrm{C}$
3. Initial temperature $\left({\mathrm{T}}_{\mathrm{i}}\right)={90}^{\mathrm{o}}\mathrm{C}$and${70}^{\mathrm{o}}\mathrm{C}$

The heat capacity is amount of heat required to raise the mass of given substance by unit degree. If the substance has unit mass, then heat capacity is called as specific heat capacity.

We can use the formula for specific heat capacity and latent heat of fusion. Using the concept of equilibrium when there is no exchange of heat and if the temperature is the same, we can find the final temperature of the mixture and remaining mass of the ice.

Formula:

The formula of mixture temperature after the whole heat absorbing and releasing process, ${\mathrm{T}}_{\mathrm{f}}=\frac{{\mathrm{C}}_{\mathrm{W}}{\mathrm{M}}_{\mathrm{W}}{\mathrm{T}}_{\mathrm{i}}-{\mathrm{L}}_{\mathrm{f}}{\mathrm{m}}_{\mathrm{i}}}{\left({\mathrm{m}}_{\mathrm{i}}+{\mathrm{M}}_{\mathrm{W}}\right){\mathrm{C}}_{\mathrm{W}}}$ …(¾±)

The formula of mass of the ice that melts after the whole heat process of melting, data-custom-editor="chemistry" ${\mathrm{M}}_{\mathrm{m}}=\frac{{\mathrm{C}}_{\mathrm{W}}{\mathrm{M}}_{\mathrm{W}}\left({\mathrm{T}}_{\mathrm{i}}-0^{\mathrm{o}}\mathrm{C}\right)}{{\mathrm{L}}_{\mathrm{f}}}$ …(¾±¾±)

Here, T_f is final temperature of water, C_W is specific heat capacity of water, T_i is initial temperature of water, L_f is latent heat of fusion of ice, m_i is mass of ice, M_W is mass of water.

It is given that the person makes a quantity of iced tea by mixing 500 g of hot tea with an equal mass of ice at its melting point. There is no exchange of energy with the environment, so we have to find the mixture temperature and the remaining mass when the initial temperature is 90^oC and 70^oC.

The system is insulated so

So, the mixture at temperature by using equation (i) is given as:

$\begin{array}{rcl}{\mathrm{T}}_{\mathrm{f}}& =& \frac{\left(4190\mathrm{J}/{\mathrm{kg}}^{\mathrm{o}}\mathrm{C}×0.500\mathrm{kg}×{90}^{\mathrm{o}}\mathrm{C}\right)-333×{10}^3\mathrm{J}/\mathrm{kg}×0.500\mathrm{kg}}{\left(0.500\mathrm{kg}+0.500\mathrm{kg}\right)4190\mathrm{J}/{\mathrm{kg}}^{\mathrm{o}}\mathrm{C}}\\ & =& 5.3^{\mathrm{o}}\mathrm{C}\end{array}$

Hence, the temperature of the mixture is data-custom-editor="chemistry" $5.3^{\mathrm{o}}\mathrm{C}$.

The remaining mass, $\left({\mathrm{M}}_{\mathrm{f}}\right)=0$

Since no ice remains at temperature ${\mathrm{T}}_{\mathrm{f}}=5.3^{\mathrm{o}}\mathrm{C}$.

If we use the formula of equation (i), we get the temperature of the mixture as given:

$\begin{array}{rcl}{\mathrm{T}}_{\mathrm{f}}& =& \frac{4190\mathrm{J}/{\mathrm{kg}}^{\mathrm{o}}\mathrm{C}×0.500\mathrm{kg}×{70}^{\mathrm{o}}\mathrm{C}-333×{10}^3\mathrm{J}/\mathrm{kg}×0.500\mathrm{kg}}{\left(0.500\mathrm{kg}+0.500\mathrm{kg}\right)4190\mathrm{J}/{\mathrm{kg}}^{\mathrm{o}}\mathrm{C}}\\ & =& -4.{73}^{\mathrm{o}}\mathrm{C}\\ & <& 0^{\mathrm{o}}\mathrm{C}\\ & & \end{array}$

This is not possible. It means that not all the ice has melted in this case, and the equilibrium temperature is data-custom-editor="chemistry" $0^{\mathrm{o}}\mathrm{C}$

The amount of ice that melts is given by the equation (ii) as:

$\begin{array}{rcl}{\mathrm{M}}_{\mathrm{m}}& =& \frac{\left(4190\mathrm{J}/{\mathrm{kg}}^{\mathrm{o}}\mathrm{C}\right)\left(0.500\mathrm{kg}\right)\left({70}^{\mathrm{o}}\mathrm{C}\right)}{333×{10}^3\mathrm{J}/\mathrm{kg}}\\ & =& 0.440\mathrm{kg}\end{array}$

Therefore, the amount of solid ice remained is,

$\begin{array}{rcl}{\mathrm{M}}_{\mathrm{f}}& =& {\mathrm{m}}_{\mathrm{i}}-{\mathrm{M}}_{\mathrm{m}}\\ & =& 0.500\mathrm{kg}-0.440\mathrm{kg}\\ & =& 0.060\mathrm{kg}\end{array}$

Hence, the value of the remaining mass is $0.06\mathrm{kg}$.