Questions
Certain quantity of water cools from 700C to 600C in the first 5 min and to 54°C in the next 5 min. The
temperature of the surrounding is
a
450C
b
200C
c
420C
d
100C
detailed solution
Correct option is A
Let temperature of surrounding be Q0 then from Newton's law of cooling Case I Q2−Q1t=KQ2+Q12−Q070−605=K65−Q0 …(i) Case II 60−545=57−Q0K …(ii)On dividing Eq. (i) by Eq. (ii), we get106=65−Q057−Q0⇒Q0=45∘CTalk to our academic expert!
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A calorimeter of mass 0.2 kg and specific heat 900 Containing 0.5 kg of a liquid of specific heat 2400J/kg/K. Its temperature falls from
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