The coil of a calorimeter C has a resistance of R1 = 60Ω. The coil R1 is connected to the circuit as shown in figure. How many degrees (°C) will 240 grammes of water poured into the calorimeter be heated in 7 minutes during which a current flows through the coil if the ammeter shows 3A? The resistance R2 = 30Ω. Disregard the resistances of the battery and the ammeter, and the heat losses and heat capacity of the calorimeter and the resistor.

Engineering

Physics

Ohm Law and Current Basics

Question

The coil of a calorimeter C has a resistance of R₁ = 60Ω. The coil R₁ is connected to the circuit as shown in figure. How many degrees (°C) will 240 grammes of water poured into the calorimeter be heated in 7 minutes during which a current flows through the coil if the ammeter shows 3A? The resistance R₂ = 30Ω. Disregard the resistances of the battery and the ammeter, and the heat losses and heat capacity of the calorimeter and the resistor.

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Current through R₁ = = 14 mg
Total heat loss = (1)² × 60 × t

$so (1)^{2} \times 60 \times t = (\frac{240 gm}{1000}) \times (4200) ΔT$

ΔT = 25°C

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