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Question: Ice point and steam point on a particular scale reads 10<sup>0</sup> and 80<sup>0</sup> respectively...

Ice point and steam point on a particular scale reads 100 and 800 respectively. The temperature on 0F scale when temperature on new scale is 450 is –

A

500 F

B

1120F

C

1220F

D

1380F

Answer

1220F

Explanation

Solution

Relation between the two scales

t108010\frac{t–10}{80–10} = F32180\frac{F–32}{180}

F = 187\frac{18}{7} (t – 10) + 32