Question

On a temperature scale Y, water freezes at $-{160}^o$ Y and boils at $-{50}^o$ Y. On this Y scale, a temperature of $340K$ is :

• A. $-{106.3}^o$ Y
• B. $-{96.3}^o$ Y
• C. $-{86.3}^o$ Y
• D. $-{76.3}^o$ Y

Answer 1

Answer: (C)

$-{86.3}^o$ Y

Since the temperature scale is assumed to be linear, slope in two cases will be same. Hence,

$\frac{Y-(-160)}{-50-(-160)}=\frac{K-273}{373-273}$,

$\frac{Y+160}{110}=\frac{K-273}{100}$

$Y=\frac{11}{10}(K-273)-160$

$Y=\frac{11}{10}(340-273)-160=-{86.3}^{0}Y$