Question

A student wants to cool 0.25kg of Coke drink (mostly water), initially 25°C, by adding ice initially at -20°C. How much ice should be added so that the final temperature will be 0°C with all the ice melted if the heat capacity of the container may be neglected.

Answer 1

Answer on Question #60843, Physics – Molecular Physics | Thermodynamics

A student wants to cool $0.25\mathrm{kg}$ of Coke drink (mostly water), initially $25{}^{\circ}\mathrm{C}$, by adding ice initially at $-20{}^{\circ}\mathrm{C}$. How much ice should be added so that the final temperature will be $0{}^{\circ}\mathrm{C}$ with all the ice melted if the heat capacity of the container may be neglected.

Solution:

Specific heat capacity, ice: $c_{\text{ice}} = 2.108 \, \text{kJ/kg} \cdot \text{K}$

Specific heat capacity, water: $c_{\text{water}} = 4.187 \, \text{kJ/kg} \cdot \text{K}$

The heat of fusion (or specific enthalpy of fusion) of ice is $L = 334 \, \text{kJ/kg}$.

After mixing, the hot liquid has cooled to a temperature $T_c = 0{}^{\circ}\mathrm{C}$.

The quantity of heat from first liquid:

$$Q_1 = c_{\text{water}} m_{\text{water}} (T_1 - T_c)$$

The energy to heat up the ice is the sum of the following

$$Q = c_{\text{ice}} m_{\text{ice}} (T_C - T_2) + L m_{\text{ice}}$$

Since heat does not disappear, and transferred from one liquid to another:

$$Q_1 = Q_2$$ $$c_{\text{water}} m_{\text{water}} (T_1 - T_c) = c_{\text{ice}} m_{\text{ice}} (T_C - T_2) + L m_{\text{ice}}$$

Thus,

$$m_{\text{ice}} = \frac{c_{\text{water}} m_{\text{water}} (T_1 - T_c)}{c_{\text{ice}} (T_C - T_2) + L}$$ $$m_{\text{ice}} = \frac{4187 \cdot 0.25 \cdot (25 - 0)}{2108 \cdot (0 + 20) + 334000} = 0.0696 \, \text{kg} \approx 0.07 \, \text{kg}$$

Answer: