Question

On a linear temperature scale y, water freeze 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)

Answer 1

Hello dear,

● Answer- 86.3 Y

● Explaination-
# Given-
Boiling point = 50 Y = 373 K
Freezing point = 160 Y = 273 K

# Solution-
Let
y = ka + b

From given data,
50 = 373a + b
160 = 273a + b

Solving this,
a = -1.1
b = 460.3

For k = 340 K
y = -1.1(340) + 460.3
y = -374 + 460.3
y = 86.3 Y

340 K will be read as 86.3 Y on new scale.

Hope that helps you...