Question 948: Thermodynamics & KTG

Question

On an imaginary linear scale of temperature (called 'W' scale) the freezing and boiling points of water are 39$$^\circ$$ W and 239$$^\circ$$ W respectively. The temperature on the new scale corresponding to 39$$^\circ$$C temperature on Celsius scale will be

Accepted Answer

In Celsius scale the freezing and boiling points are 0$$^\circ$$C and 100$$^\circ$$C. In the given imaginary scale the freezing and boiling points are 39$$^\circ$$ W and 239$$^\circ$$ W. Hnece we can write
$$\mathrm{\frac{C}{100}=\frac{W-39}{200}}$$
For C = 39$$^\circ$$, $$\frac{39}{100}=\mathrm{\frac{W-39}{200}}$$
Solving, W = 117$$^\circ$$C

Answer

139$$^\circ$$ W

Answer

78$$^\circ$$ W

Answer

117$$^\circ$$ W

Answer

200$$^\circ$$ W

Thermodynamics & KTG

On an imaginary linear scale of temperature (called 'W' scale) the freezing and boiling points of water are 39$^\circ W and 239^\circ W respectively. The temperature on the new scale corresponding to 39^\circ$C temperature on Celsius scale will be

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