Question

How many grams of ice must we add in 500g of water which has temperature 30 °C so when the ice melts completely, the final temperature of water would be 5 °C?

Answer 1

Answer on question #61569, Physics / Molecular Physics | Thermodynamics

Statement of a problem

How many grams of ice must we add in 500g of water which has temperature $30{}^{\circ}\mathrm{C}$ so when the ice melts completely, the final temperature of water would be $5{}^{\circ}\mathrm{C}$?

$t_1 = 30{}^{\circ}C$ – initial temperature of the water;

$m_w = 0.5\,kg$ – mass of the water;

$t_2 = 5{}^{\circ}C$ – final temperature of water;

$t_{ice} = 0{}^{\circ}C$ – temperature of the ice;

$\lambda = 3.5 \times 10^{5}J/kg$ – specific heat of fusion of ice

$C = 4.2 \times 10^{3}J/kg$ – specific heat capacity of water

Find: $m_{ice}$ – mass of the ice

Thermal balance equation

$$Q_f + Q_1 = Q_2,$$

Where $Q_f = \lambda m_{ice}$ – heat of fusion,

$Q_1 = C m_{ice}(t_2 - t_{ice})$ – heat expended on heating the melt water

$Q_2 = C m_w(t_1 - t_2)$ – heat from hot water

$$\lambda m_{ice} + C m_{ice}(t_2 - t_{ice}) = C m_w(t_1 - t_2)$$ $$m_{ice}(\lambda + C(t_2 - t_{ice})) = C m_w(t_1 - t_2)$$ $$m_{ice} = \frac{C m_w(t_1 - t_2)}{\lambda + C(t_2 - t_{ice})}$$

Calculation:

$$m_{ice} = \frac{4.2 \times 10^{3} \times 0.5 \times (30 - 5)}{3.3 \times 10^{4} + 4.2 \times 10^{3} \times (5 - 0)} = \frac{2.1 \times 25 \times 10^{3}}{3.3 \times 10^{4} + 2.1 \times 10^{4}} = \frac{5.25}{35.1} \approx 0.1496\,(kg)$$ $$m_{ice} \approx 0.1496\,kg = 149.6\,g$$

Answer: $m_{ice} \approx 0.1496\,kg = 149.6\,g$

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