Question

How many kilograms of ice at a temperature of − 21.0 ∘C must be dropped in the water to make the final temperature of the system 23.3 ∘C?

Answer 1

Answer:

0.1826 kg of ice is to be dropped in water to make the temperature of system to 23.3 C.

Explanation:

The amount of the heat lost by the water is the heat gained by the ice so by applying the balance of heat introduced and removed from the system, (assuming the container is insulated and is of negligible mass)

$m_ic_i\Delta T_i+m_ic_w\Delta T_2+m_i L_f=m_wc_w\Delta T_3$

Here

• $m_i$ is the mass of ice which is to be calculated.
• $c_i$ is the specific heat of ice which is 2100 J/kg.K

• $c_w$ is the specific heat of ice which is 4190 J/kg.K
• $\Delta T_i$ is the change of temperature from -21 C to 0 C.
• $\Delta T_2$ is the change of temperature from 0 to 23.3 C.
• $L_f$ is the Latent heat of Fusion which is 3.34×105 J/kg
• $m_w$ is the mass of water which is given as 0.345 kg.
• $\Delta T_3$ is the change of temperature from 83.4 C to 23.3 C.

                              $m_i=\frac{m_wc_w\Delta T_3}{c_i\Delta T_i+c_w\Delta T_2+L_f}\\m_i=\frac{0.345 \times 4190 (83.4-23.3)}{2100 \times (0-(-21.0))+4190 \times (23.3-0)+3.34 \times 10^5}\\m_i=0.1826 \, kg$

Here 0.1826 kg of ice is to be dropped in water.