Problem — From power to specific heat

Problem

A heater delivers constant power $P = 800$ W. Immersed in $2.0$ kg of an unknown liquid substance initially at $15\ ^{\circ}$C, in $5$ minutes it brings the liquid to $45\ ^{\circ}$C. Determine the specific heat, assuming negligible losses. What liquid might it be?

Solution

Energy supplied. In $t = 5$ min $= 300$ s the heater delivers $Q = P \, t = 800 \cdot 300 = 2.4 \cdot 10^{5}\ \mathrm{J}$ Specific heat. From $Q = m \, c \, \Delta T$, with $m = 2.0$ kg and $\Delta T = 45 - 15 = 30$ K: $c = \frac{Q}{m \, \Delta T} = \frac{2.4 \cdot 10^{5}}{2.0 \cdot 30} = 4000\ \mathrm{J/(kg \cdot K)}$ Identification. The value found is close to that of water ($4186$): the liquid is probably a water-based substance, such as water itself or milk. $\ev{c = 4000\ \mathrm{J/(kg \cdot K)}}$

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Topics: Termologia Concepts: Calore specifico e capacità termica