A metal bowl of mass 700 g contains 1 litre of water at 27 °C. When a steel ball of mass 300 g at 120 °C is dropped into the water in the bowl, the final temperature of water is measured to be 30 °C. Specific heat capacities of steel and water are 500 J kg⁻¹ K⁻¹ and 4200 J kg⁻¹ K⁻¹, respectively. Out of the metals given in the table, what could be the metal that the bowl is made of?

2019 A/L Physics Past Paper · Question 42 (English Medium)

The metal of the bowl from a heat exchange

A metal bowl of mass $700\ \text{g}$ contains $1$ litre of water at $27\ ^\circ\text{C}$ . When a steel ball of mass $300\ \text{g}$ at $120\ ^\circ\text{C}$ is dropped into the water in the bowl, the final temperature of water is measured to be $30\ ^\circ\text{C}$ . Specific heat capacities of steel and water are $500\ \text{J kg}^{-1}\text{ K}^{-1}$ and $4200\ \text{J kg}^{-1}\text{ K}^{-1}$ , respectively. Out of the metals given in the table, what could be the metal that the bowl is made of?

(1) Aluminium
(2) Copper
(3) Lead
(4) Iron
(5) Silver

Answer and explanation

Answer: (2) Copper

What the question asks

We know:

The ball cools from 120 °C to 30 °C.

The water and the bowl warm from 27 °C to 30 °C.

We want to find which metal in the table the bowl is made of.

Relevant theory

The principle of heat exchange.

The heat lost by the hot ball equals the heat gained by the water and the bowl, plus any heat lost to the air.

The heat change of a body is $Q = mc\Delta\theta$ .

The loss to the air can be zero or more. It can never be less than zero.

According to the principle of heat exchange; If the heat loss to the environment is negligible Heat removed from the hot object = Heat acquired by the cold object

Required equations

$Q = mc\,\Delta\theta$

$Q_{ball} = Q_{water} + Q_{cont} + Q_{loss},\quad Q_{loss} \ge 0$

Reading the options

$13500$

$Q_{ball} = 0.3 \times 500 \times (120 - 30)$

$12600$

$Q_{water} = 1 \times 4200 \times (30 - 27)$

$900$

$Q_{cont} \le 13500 - 12600$

Reasoning and calculation

The bowl takes $0.7 \times c \times 3$ J.

This cannot be more than $900$ J.

$c \le \dfrac{900}{2.1} \approx 429\ \text{J kg}^{-1}\text{ K}^{-1}$

It is equal only when nothing is lost to the air.

Which metals have $c \le 429$ ?

Copper ( $385$ ), silver ( $230$ ) and lead ( $128$ ).

Aluminium ( $900$ ) and iron ( $450$ ) are too big.

$808.5$

$0.7 \times 385 \times 3$

$91.5$

$900 - 808.5$

$417$

$900 - 0.7 \times 230 \times 3$

$631$

$900 - 0.7 \times 128 \times 3$

Copper fits with a small loss, about $90$ J.

Silver and lead would need most of the heat to leak into the air.

So the bowl is copper.

This metal needs $0.7 \times 900 \times 3 = 1890$ J. Only $900$ J is left.

This metal needs only $269$ J. Then $631$ J would leak into the air. That is not believable.

This metal needs $0.7 \times 450 \times 3 = 945$ J. Only $900$ J is left.

This metal needs only $483$ J. Then $417$ J would leak into the air. That is a big loss.

Checks

Check.

With no loss, $c = 429$ . This is nearest to iron ( $450$ ).

But iron is not possible. It needs more heat than there is.

That would make energy from nothing.

The loss to the air is never negative.

$(0.3 * 500 * 90 - 4200 * 3) / (0.7 * 3)$

Conclusion

At most $900$ J is left for the bowl, so $c \le 429$ .

Iron and aluminium are not possible.

Copper fits with a small loss.

The answer is option (2).

Working out $c = 429$ and choosing the nearest value, $450$ (iron).

Iron needs $945$ J. That is more than the heat available. It would break the conservation of energy. The loss to the air is never negative.

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