Problem — Four cups

Problem

Four cups each contain $200$ g of a different substance, all at $20\ ^{\circ}$C. Each is given the same amount of heat $Q = 5000$ J. The substances are:

Cup	Substance	$c$ [J/(kg·K)]
A	water	4186
B	olive oil	1970
C	powdered aluminium	900
D	silica sand	830

Order the cups by final temperature, from coldest to hottest.

Solution

Temperature rise. From $Q = m \, c \, \Delta T$ we get $\Delta T = Q / (m \, c)$, with $m = 0.2$ kg the same for all: $\Delta T_A = \frac{5000}{0.2 \cdot 4186} \approx 5.97\ \mathrm{K}, \quad \Delta T_B = \frac{5000}{0.2 \cdot 1970} \approx 12.69\ \mathrm{K}$ $\Delta T_C = \frac{5000}{0.2 \cdot 900} \approx 27.78\ \mathrm{K}, \quad \Delta T_D = \frac{5000}{0.2 \cdot 830} \approx 30.12\ \mathrm{K}$ Final temperatures. Adding the initial $20\ ^{\circ}$C gives $T_A \approx 26.0\ ^{\circ}$C, $T_B \approx 32.7\ ^{\circ}$C, $T_C \approx 47.8\ ^{\circ}$C, $T_D \approx 50.1\ ^{\circ}$C. Conclusion. For the same heat and mass, the smaller the specific heat $c$, the more the substance heats up. Water, with the highest $c$, stays the coldest; sand, with the lowest $c$, becomes the hottest. $\ev{A < B < C < D}$

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Topics: Termologia Concepts: Calore specifico e capacità termica