On a linear temperature scale Y, water freezes at -160° Y and boils at -50° Y. On this Y scale, a temperature of 340 K would be read as: (water freezes at 273 K and boils at 373 K)

Question

On a linear temperature scale Y, water freezes at -160° Y and boils at -50° Y. On this Y scale, a temperature of 340 K would be read as: (water freezes at 273 K and boils at 373 K) (A) -73.7° Y (B) -233.7° Y (C) -86.3° Y (D) -106.3° Y

Tardigrade Admin · Accepted Answer

(Reading on any scale - LFP/UFP-LFP) = constant for all scales (340-273/373-273) = (°y-(-160)/-50-(-160)) ⇒ (67/100) = (y + 160 /110) ∴ y=-86.3°y

Q. On a linear temperature scale $Y$ , water freezes at $- 16 0^{\circ}$ $Y$ and boils at $- 5 0^{\circ} Y$ . On this $Y$ scale, a temperature of $340 K$ would be read as : (water freezes at $273 K$ and boils at $373 K$ )

$- 73. 7^{\circ} Y$

$- 233. 7^{\circ} Y$

$- 86. 3^{\circ} Y$

$- 106. 3^{\circ} Y$

$\frac{R e a d in g o n an y sc a l e - L FP}{U FP - L FP}$
= constant for all scales
$\frac{340 - 273}{373 - 273} = \frac{^{\circ} y - ( - 160 )}{- 50 - ( - 160 )}$
$\Rightarrow \frac{67}{100} = \frac{y + 160}{110}$
$∴ y = - 86. 3^{\circ} y$

Q. On a linear temperature scale Y, water freezes at −160∘ Y and boils at −50∘Y. On this Y scale, a temperature of 340K would be read as : (water freezes at 273K and boils at 373K)

−73.7∘Y

−233.7∘Y

−86.3∘Y

−106.3∘Y