Question 506: Thermodynamics & KTG

Question

A pendulum clock loses $$12$$ $$s$$ a day if the temperature is $${40^ \circ }C$$ and gains $$4$$ $$s$$ a day if the temperature is $${20^ \circ }C.$$ The temperature at which the clock will show correct time, and the co-efficient of linear expansion $$\left( \alpha  \right)$$  of the metal of the pendulum shaft are respectively :

Accepted Answer

Time lost/gained per day $$ = {1 \over 2} \propto \Delta 	heta  	imes 86400$$ second
$$12 = {1 \over 2}\alpha \left( {40 - 	heta } ight) 	imes 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i ight)$$
$$4 = {1 \over 2}\alpha \left( {	heta  - 20} ight) 	imes 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} ight)$$
On dividing we get, $$\,\,\,3 = {{40 - 	heta } \over {	heta  - 20}}$$
$$3	heta  - 60 = 40 - 	heta $$
$$4	heta  = 100 \Rightarrow 	heta  = {25^ \circ }C$$

Answer

$${30^ \circ }C;\,\,\alpha  = 1.85 	imes {10^{ - 3}}/{}^ \circ C$$

Answer

$${55^ \circ }C;\,\,\alpha  = 1.85 	imes {10^{ - 2}}/{}^ \circ C$$

Answer

$${25^ \circ }C;\,\,\alpha  = 1.85 	imes {10^{ - 5}}/{}^ \circ C$$

Answer

$${60^ \circ }C;\,\,\alpha  = 1.85 	imes {10^{ - 4}}/{}^ \circ C$$

Thermodynamics & KTG

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