An engineering laboratory is testing a new low-melting-point metal alloy, alloy M, designed for use inside the fusible links of automatic fire-sprinkler systems. Alloy M must melt reliably at a known, fixed temperature so that a sprinkler activates as soon as a fire starts. A te…

Answer: ((a)) 60 °C; identified from the plateau (constant temperature) in the data between t=4 min and t=7 min, where the temperature stops rising while the alloy changes from solid to liquid. ((b)) The energy supplied by the furnace is used to overcome (weaken) the forces of attraction holding the particles in their fixed positions, allowing them to move past one another, rather than increasing the particles' average kinetic energy. Since temperature is a measure of the av… ((c)) 60-20/4-0 = 40/4 = 10 °C/min ((d)) The plateau would last longer. A larger mass contains more particles overall, so more total energy is needed to overcome the forces of attraction between all of them; since energy is supplied at the same constant rate as before, this takes more time. The melting point itself wou…. Part (a): Identifying the melting point Reading down the temperature row, the value stops increasing and stays constant at 60 °C for four consecutive readings, from t=4 min to t=7 min. This flat section (plateau) on the temperature–time da…

IGCSE Chemistry 0620: Interpreting a Heating Curve with Kinetic Particle Theory

Time / min	0	1	2	3	4	5	6	7	8	9
Temperature / $^{\circ}\text{C}$	20	30	40	50	60	60	60	60	70	80

Time / min

Temperature /

^{\circ}\text{C}

Part (a): Identifying the melting point

Reading down the temperature row, the value stops increasing and stays constant at $60\,^{\circ}\text{C}$ for four consecutive readings, from $t=4$ min to $t=7$ min. This flat section (plateau) on the temperature–time data is where the solid alloy is turning into a liquid. Both states are present together at a single, fixed temperature.

Melting point of alloy M $= \boxed{60\,^{\circ}\text{C}}$ , identified from the plateau in the recorded temperatures.

Part (b): Why the temperature stays constant during melting

Before melting starts, the particles in solid alloy M are held in fixed positions by strong forces of attraction, only vibrating about those positions. As the furnace continues to supply energy at a constant rate between $t=4$ min and $t=7$ min, this energy is used to overcome (weaken) the forces of attraction between the particles, freeing them to move past one another as a liquid, rather than to speed the particles up.

Because temperature is a measure of the average kinetic energy of the particles, and this average kinetic energy is not increasing during melting, the temperature stays constant even though energy is still being transferred into the alloy throughout this time.

Part (c): Rate of temperature increase before melting

Between $t=0$ and $t=4$ minutes, the alloy is solid and its temperature rises steadily from $20\,^{\circ}\text{C}$ to $60\,^{\circ}\text{C}$ :

$\text{rate} = \frac{\text{change in temperature}}{\text{change in time}} = \frac{60 - 20}{4 - 0} = \frac{40}{4} = 10\,^{\circ}\text{C/min}$

Part (d): Effect of a larger mass on the plateau

Melting a larger mass of alloy M means there are more particles overall, and every one of them needs its forces of attraction overcome before it can move as a liquid. This means a greater total amount of energy must be supplied to melt the whole sample.

Since the furnace supplies energy at the same constant rate as before, supplying more total energy takes more time, so the plateau (constant-temperature section) of the new graph would be longer.

The temperature at which the plateau occurs would not change: it would still be $60\,^{\circ}\text{C}$ , because the melting point is a property of the substance (alloy M) itself, not of how much of it is being heated.

Final answers

(a) Melting point $= \boxed{60\,^{\circ}\text{C}}$ , read from the plateau between $t=4$ min and $t=7$ min.
(b) Energy overcomes the forces between particles instead of increasing their kinetic energy, so temperature (average kinetic energy) stays constant.
(c) Rate $= \boxed{10\,^{\circ}\text{C/min}}$ .
(d) The plateau would be longer (more particles need more total energy, supplied at the same rate), but it would still occur at $60\,^{\circ}\text{C}$ .