First slide

Temperature and heat-thermal properties of matter

Question

If a thermometer reads freezing point of water as $20^{\circ} c$ and boiling point is $150^{\circ} c$ how much thermometer read, when the actual temperature is $60^{\circ} c$

Moderate

A

$98^{\circ} c$
B

$110^{\circ} c$
C

$40^{\circ} c$
D

$60^{\circ} c$

Solution

$\begin{array}{rcl} u n k n o w n t h e r m o m e t e r, L o w e r F i x e d p o i n t (LFP) & = & 20^{0} C; U p p e r F i x e d P o i n t (UFP) = 150^{0} C; a c t u a l r e a d i n g g i v e n = 60^{0} C \\ t h e f o r m u l a t h a t r e l a t e s u n k n o w n t h e r m o m e t e r a n d c e l s i u s t h e r m o m e t e r i s, \\ \frac{Z-LFP}{UFP-LFP} & = & \frac{C-LFP}{UFP-LFP} \\ c e l s i u s t h e r m o m e t e r h a s LFP & = & 0^{0} C a n d U F P = 100^{0} C \\ s u b s t i t u t e g i v e n v a l u e s i n a b o v e e q u a t i o n, x i s r e a d i n g o n u n k n o w n t h e r m o m e t e r \\ \frac{x-20}{150-20} = \frac{60-0}{100-0} \\ x-20=78 \\ x= 98^{\circ} c \end{array}$

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