Question

A particular temperature scale is obtained according to the relation: \[ t = a e^{\alpha} + b \] where \(a\) and \(b\) are constants, and \(t\) is in °C. The thermometric property as measured by the thermometer is \(\alpha\). The values of \(\alpha\) at ice point and steam point are 6 and 9, respectively. The temperature (in °C) which gives \(\alpha = 7\) is ................. (rounded off to two decimal places).

Hint:

Always calibrate thermometer scales using ice and steam points. Then use exponential interpolation for intermediate values.

Answer 1

Correct Answer: 8.9

Step 1: Write given relation.
\[ t = a e^{\alpha} + b \]

Step 2: Use calibration conditions.
At ice point (\(t = 0^\circ C, \alpha = 6\)): \[ 0 = a e^6 + b \Rightarrow b = -a e^6 \] At steam point (\(t = 100^\circ C, \alpha = 9\)): \[ 100 = a e^9 + b \] Substitute \(b = -a e^6\): \[ 100 = a e^9 - a e^6 \] \[ a (e^9 - e^6) = 100 \] \[ a = \frac{100}{e^9 - e^6} \]

Step 3: Find temperature at \(\alpha = 7\).
\[ t = a e^7 + b \] \[ = a e^7 - a e^6 \] \[ = a (e^7 - e^6) \] \[ = \frac{100 (e^7 - e^6)}{e^9 - e^6} \] Numerical calculation: \[ e^6 = 403.43, e^7 = 1096.63, e^9 = 8103.08 \] \[ t = \frac{100 (1096.63 - 403.43)}{8103.08 - 403.43} \] \[ t = \frac{100 (693.20)}{7699.65} = 8.99 \] Oops! Let’s carefully check again: Wait! Recalculate: \[ t = \frac{100 (e^7 - e^6)}{e^9 - e^6} \] Numerator: \(e^7 - e^6 = 1096.63 - 403.43 = 693.20\). Denominator: \(e^9 - e^6 = 8103.08 - 403.43 = 7699.65\). \[ t = \frac{100 \times 693.20}{7699.65} = 8.99^\circ C \] Final Answer:
\[ \boxed{8.99^\circ C} \]