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The 2.0Ω resistor shown in figure (33-E1) is dipped into a calorimeter containing water. The heat capacity of the calorimeter together with water is 2000J/K. (a) If the circuit is active for 15minutes, what would be the rise in the temperature of the water? (b) Suppose the 6.0Ω resistor gets burnt. What would be the rise in the temperature of the water in the next 15minutes?

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Reff = 12/8 + 1 = 5/2

i = 6/(5/2) = 12/5Amp.

2000J of heat raises the temp. by 1K

5832J of heat raises the temp. by 2.916K.

(b) When 6Ω resistor get burnt Reff = 1 + 2 = 3Ω

i = 6/3 = 2Amp.

Heat = 2 × 2 × 2 ×15 × 60 = 7200J

2000J raises the temp. by 1K

7200J raises the temp by 3.6k

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