Question

The drawing shows two thermometers, A and B, whose
temperatures are measured in °A and °B. The ice and boiling
points of water are also indicated.
(a) Using the data in the drawing, determine the number of B
degrees on the B scale that correspond to 1 A° on the A scale.
(b) If the temperature of a substance reads +40.0°A on the A
scale, what would that temperature read on the B scale?
(Cutnell 354)

Answer 1

a) 1 A° corresponds to 1.22 B°

b) $105.4^{\circ}B$

Explanation:

a)

The image of the problem is missing: find it in attachment.

Here we have two temperature scales:

- In scale A, the freezing point is at $-30.0^{\circ}A$, while the boiling point is at a temperature of $+60.0^{\circ}A$

- In scale B, the freezing point is at $+20.0^{\circ}B$, while the boiling point is at a temperature of $+130.0^{\circ}B$

The freezing point and the boiling point must be at the same absolute temperature on the two scales: this means that the interval

(boiling point - freezing point)

must be the same in the two scales.

For scale A, this interval is equal to:

$\Delta T_A = +60.0-(-30.0)=90.0^{\circ}A$

For scale B, this interval is equal to:

$\Delta T_B=+130-(+20)=110^{\circ}B$

Therefore, 1 degree on the A scale corresponds to:

$\frac{x^{\circ}B}{1^{\circ}A}=\frac{110^{\circ}B}{90.0^{\circ}A}\\x^{\circ}B=\frac{110}{90}\cdot 1=1.22$

So, 1 A° corresponds to 1.22 B°.

b)

The temperature of the substance on the A scale is

$T_A=+40.0^{\circ}A$

This means that the difference of its temperature relative to the freezing point on the A scale is

$\Delta T_A=+40.0-(-30.0)=70.0^{\circ}A$

In part A, we stated that 1 degree on the A scale corresponds to 1.22 degrees on the B scale, so this temperature interval on the B scale is

$\Delta T_B=1.22\Delta T_A=1.22(70)=85.4^{\circ}B$

And therefore, the temperature of the substance on the B-scale is equal to the freezing point (on the B scale) plus this interval:

$T_B=+20.0+85.4=105.4^{\circ}B$