Topic: Thermodynamics
A mercury thermometer was first placed in melting ice and the length of mercury column was observed to be 10 mm; when it was placed in steam, the length of the column was 250 mm. When placed in tap water, the length of the column was 58 mm. The temperature of the tap water is
- A. 24.2°C
- B. 20°C
- C. 38.4°C
- D. 4.14°C
Answer: B. 20°C
Explanation: Assuming a linear temperature profile, the equation is $T = al + b$. For melting ice ($0^{\circ}C$, $l = 10 mm$), $0 = 10a + b$. For steam ($100^{\circ}C$, $l = 250 mm$), $100 = 250a + b$. Subtracting the first equation from the second gives $100 = 240a$, leading to $a = \frac{5}{12}$. Substituting $a$ back into the first equation yields $b = -10 \times \frac{5}{12} = -\frac{25}{6}$. The temperature equation is $T = \frac{5}{12}l - \frac{25}{6}$. For tap water ($l = 58 mm$), $T = \frac{5}{12}(58) - \frac{25}{6} = 20^{\circ}C$.