Q. 12

The 2.0 Ω resistor shown in figure is dipped into a calorimeter containing water. The heat capacity of the calorimeter together with water is 2000 J K–1.

(a) If the circuit is active for 15 minutes, what would be the rise in the temperature of the water?

(b) Suppose the 6.0 Ω resistor gets burnt. What would be the rise in the temperature of the water in the next 15 minutes?



Answer :

Given-


Resistance - 2.0 ΩΩ


heat capacity of the calorimeter together with water = 2000 J K–1



Looking into the circuit the effective resistance of the circuit,


2.0 Ω in parallel with 6.0 Ω and this combination in series with 1.0 Ω




Now, current i through the circuit, from ohm’s law-






Let take i’ as the current through the 6 Ω resistor.


Then, applying KCL, the algebraic sum of current entering and leaving a node is zero and from ohm’s law, we can write -









(a) Heat generated in the 2 Ω resistor, using Joule’s Heating effect




from (1) and substituting values




Given that the heat capacity of the calorimeter together with water is 2000 J K−1


Which means , 2000 J of heat raise the temp by 1 K.



Then, 5832 J of heat raises the temperature of water by




(b) When the 6 Ω resistor burn out, the effective resistance of the


circuit will become –


Untitled.png





Current through the circuit,





Heat generated in the 2 Ω resistor,
using Joule’s Heating effect





Given that the heat capacity of the calorimeter together with water is 2000 J K−1


Which means , 2000 J of heat raise the temp by 1 K.



So, 7200J will raise the temperature by



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