2020 A/L Physics Past Paper · Question 35 (English Medium)

Answer: (5) $400\ \text{g}$

Given information

Given: water $m_w = 1$ kg, $35^\circ\text{C}$, $c_w = 4.0 \times 10^3\ \text{J kg}^{-1}\,{}^\circ\text{C}^{-1}$; ice at $-5^\circ\text{C}$, $c_i = 2.0 \times 10^3$; $l = 3.4 \times 10^5\ \text{J kg}^{-1}$; no heat exchanged with the container or surroundings.

What the question asks

We must find: the largest mass of ice that can melt completely.

Relevant theory

The amount of heat required to raise the temperature of a unit mass of a substance by 1 unit of temperature (1 °C or 1 K) defined as the specific heat capacity of that substance.

Heat removed from the hot object = Heat acquired by the cold object

Thermal energy is absorbed by the solid for this process during the whole period in which this transformation occurs. Even though the thermal energy is absorbed there is no increase in temperature.

Specific latent heat of fusion of a solid material is defined as the heat required to change the state of a unit mass of a solid to liquid without change of temperature (at its melting point).

Required equations

$Q_w = m_w c_w (35 - 0)$

$Q_i = m c_i (0 - (-5)) + m l$

$Q_w = Q_i$

Applying it to this question

Plan: while any ice remains, the mixture cannot fall below $0^\circ\text{C}$; so the most heat the water can give is from $35 \to 0^\circ\text{C}$.

(1) that heat; (2) the heat needed per kg of ice; (3) divide.

Reasoning and calculation

$1.4 \times 10^5$

$Q_w = 1 \times 4.0 \times 10^3 \times 35$

$1.0 \times 10^4$

$2.0 \times 10^3 \times 5$

$3.4 \times 10^5$

$l$

$3.5 \times 10^5$

$1.0 \times 10^4 + 3.4 \times 10^5$

$0.40$

$m = \frac{1.4 \times 10^5}{3.5 \times 10^5}$

$400$

$m = 0.40 \times 1000$

Checks

Unit check: $\text{kg} \times \text{J kg}^{-1}\,{}^\circ\text{C}^{-1} \times {}^\circ\text{C} = \text{J}$; $\dfrac{\text{J}}{\text{J kg}^{-1}} = \text{kg}$ ✓.

Physical plausibility: most of the heat goes into melting (only $1.0 \times 10^4$ of $3.5 \times 10^5$ J warms the ice up to $0^\circ\text{C}$); $1$ kg of water at $35^\circ\text{C}$ melting $400$ g of ice is reasonable ✓.

$mw * cw * th0 / (ci * 5 + L) * 1000$

Conclusion

$m = \frac{1.4 \times 10^5}{3.5 \times 10^5} = 0.40\ \text{kg}$

The water gives $1.4 \times 10^5$ J down to $0^\circ\text{C}$; each kg of ice needs $3.5 \times 10^5$ J, melting $0.40$ kg — the answer is (5) $400$ g.

Forgetting the heat needed to warm the ice from $-5^\circ\text{C}$ to $0^\circ\text{C}$.

Before it can melt, the ice must first reach $0^\circ\text{C}$: add $m c_i \times 5$ as well.