2021 A/L Physics Past Paper · Question 42 (English Medium)

Answer: (4) $4\ \text{g}$

Given information

We know:

Copper vessel: $m_c = 2.0\ \text{kg}$, $c_c = 400\ \text{J kg}^{-1}\ \text{K}^{-1}$, starting at $150\ ^\circ\text{C}$.

Water: $m_w = 0.1\ \text{kg}$, $c_w = 4000\ \text{J kg}^{-1}\ \text{K}^{-1}$, starting at $25\ ^\circ\text{C}$.

Latent heat of vapourization: $L = 2.5 \times 10^6\ \text{J kg}^{-1}$. No heat is lost to the surroundings.

What each letter means:

$m_c$ and $m_w$ ($\text{kg}$): mass of the copper vessel and mass of the water.

$c_c$ and $c_w$ ($\text{J kg}^{-1}\ \text{K}^{-1}$): specific heat capacity, the heat needed to warm $1\ \text{kg}$ by $1\ \text{K}$.

$\theta$ ($^\circ\text{C}$): temperature.

$L$ ($\text{J kg}^{-1}$): specific latent heat of vapourization, the heat that turns $1\ \text{kg}$ of water at its boiling point into vapour.

$Q$, $Q_1$ and $Q_2$ ($\text{J}$): heat. $Q_1$ is the heat the copper gives out. $Q_2$ is the heat that warms the water to $100\ ^\circ\text{C}$.

$m$ ($\text{kg}$): mass of water that turns into vapour.

What the question asks

We want: the mass of water that turns into vapour.

The plan: some water boils, so everything ends at $100\ ^\circ\text{C}$. Find the heat the copper gives out, take away the heat that warms the water to $100\ ^\circ\text{C}$, and use what is left to boil water.

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.

If the heat loss to the environment is negligible Heat removed from the hot object = Heat acquired by the cold object

Specific latent heat of vapourization of a liquid is defined as the heat required to change the state of a unit mass of a liquid to its vapour without change of temperature (at its boiling point).

Required equations

$Q = mc\theta$

$Q = mL$

$\text{heat given by copper} = \text{heat taken by water}$

Applying it to this question

For example, Amma sprinkles a few drops of water on a hot dosa pan to see if it is ready.

The hot pan gives its heat to the drops, and the pan cools a little.

First that heat warms the drops up to $100\ ^\circ\text{C}$.

Then the heat that is left turns some drops into steam. You hear a hiss and see a puff of steam.

A little heat left over makes only a little steam. The other drops stay on the pan as hot water.

$$m = \dfrac{Q_1 - Q_2}{L}$$

In words: $$the\,mass\,of\,steam = the\,left-over\,heat\,divided\,by\,the\,heat\,that\,one\,kilogram\,needs$$

$40000$

$Q_1 = 2.0 \times 400 \times (150 - 100)$

$30000$

$Q_2 = 0.1 \times 4000 \times (100 - 25)$

$10000$

$Q_1 - Q_2 = 40000 - 30000$

$0.004$

$m = \dfrac{10000}{2.5 \times 10^6}$

$\dfrac{\text{J}}{\text{J kg}^{-1}} = \text{kg}$, so $m = 0.004\ \text{kg} = 4\ \text{g}$.

Reasoning and calculation

The copper gives $40\,000\ \text{J}$. Warming the water takes $30\,000\ \text{J}$.

The last $10\,000\ \text{J}$ turns $4\ \text{g}$ of water into vapour. The answer is (4).

Checks

$(a * b * c - d * e * f) / l$

Does it make sense?

The copper has more heat to give ($40\,000\ \text{J}$) than the water needs to reach $100\ ^\circ\text{C}$ ($30\,000\ \text{J}$). So some water must boil.

Only $4\ \text{g}$ of the $100\ \text{g}$ boils. So water is left, and $100\ ^\circ\text{C}$ is the right final temperature.

Conclusion

$m = \dfrac{m_c c_c (150 - 100) - m_w c_w (100 - 25)}{L} = 4\ \text{g}$

Heat left $= 40\,000 - 30\,000 = 10\,000\ \text{J}$.

$m = \dfrac{10\,000}{2.5 \times 10^6} = 0.004\ \text{kg} = 4\ \text{g}$. The answer is (4).

When a hot body boils some water, the final temperature is the boiling point, $100\ ^\circ\text{C}$.

Warm the water to $100\ ^\circ\text{C}$ first. Only the heat left after that boils water.

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