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 –

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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