Question

A thermometer has wrong calibrations. It reads melting point of ice as $-{10}^{o}C$ and reads ${60}^{o}C$ at temperature of ${50}^{o}C$. Temperature of boiling point noted by this thermometer will be (All temperature are note d atmospheric pressure)

• A. ${120}^{o}C$
• B. ${110}^{o}C$
• C. ${90}^{o}C$
• D. ${130}^{o}C$

Answer 1

Answer: (D)

${130}^{o}C$

$\begin{array}{c}\text{for comparision between}\\ \text{two scales we use:}\\ {\left(\frac{T-T_{mp}}{T_{bp}-T_{mp}}\right)}_1={\left(\frac{T-T_{mp}}{T_{bp}-T_{mp}}\right)}_2\end{array}$

$\begin{array}{c}\Rightarrow \frac{50-0}{100-0}=\frac{60-(-10)}{x-(-10)}\\ \Rightarrow \frac{1}{2}=\frac{70}{x+10}\\ \Rightarrow x+10=140\\ \Rightarrow x={130}^{\circ }C\end{array}$