JEE MAIN - Physics (2020 - 8th January Evening Slot - No. 24)
Three containers C1, C2 and C3 have water at
different temperatures. The table below shows
the final temperature T when different amounts
of water (given in litres) are taken from each
containers and mixed (assume no loss of heat
during the process)
_8th_January_Evening_Slot_en_24_1.png)
The value of $$\theta $$ (in °C to the nearest integer) is ..........
The value of $$\theta $$ (in °C to the nearest integer) is ..........Answer
50
Explanation
Let containers C1, C2, C3 contain water at T1, T2 and T3 temperatures respectively.
Applying law of calorimetry
1(T1 – 60) + 2(T2 – 60) = 0
$$ \Rightarrow $$ T1 + 2T2 = 180 ....(1)
1(T2 – 30) + 2(T3 – 30) = 0
$$ \Rightarrow $$ T2 + 2T3 = 90 ......(2)
2(T1 – 60) + 1(T3 – 60) = 0
$$ \Rightarrow $$ 2T1 + T3 = 180 .....(3)
from (1), (2), (3)
T1 = 80o C
T2 = 50o C
T3 = 20o C
1 × (T1 – $$\theta $$) + 1 ×(T2 – $$\theta $$) + 1(T3 – $$\theta $$) = 0
$$ \Rightarrow $$ T1 + T2 + T3 = 3$$\theta $$
$$ \Rightarrow $$ $$\theta $$ = $${{80 + 50 + 20} \over 3}$$ = 50oC
Applying law of calorimetry
1(T1 – 60) + 2(T2 – 60) = 0
$$ \Rightarrow $$ T1 + 2T2 = 180 ....(1)
1(T2 – 30) + 2(T3 – 30) = 0
$$ \Rightarrow $$ T2 + 2T3 = 90 ......(2)
2(T1 – 60) + 1(T3 – 60) = 0
$$ \Rightarrow $$ 2T1 + T3 = 180 .....(3)
from (1), (2), (3)
T1 = 80o C
T2 = 50o C
T3 = 20o C
1 × (T1 – $$\theta $$) + 1 ×(T2 – $$\theta $$) + 1(T3 – $$\theta $$) = 0
$$ \Rightarrow $$ T1 + T2 + T3 = 3$$\theta $$
$$ \Rightarrow $$ $$\theta $$ = $${{80 + 50 + 20} \over 3}$$ = 50oC
Comments (0)
