Question

[Physics- Class X]

A heating coil is immersed in a calorimeter of heat capacity $\sf 50 J {°C}^{ - 1}$ containing 1.0 kg of a liquid of specific heat capacity $\sf 450 J \: \: { {kg}^{ - 1} \: \: °C}^{ - 1}$. The temperature of liquid rises by 10°C when 2.0 A current is passed for 10 minutes. Find :
(i) the resistance of the coil,
(ii) the potential difference
across the coil.
State the assumption used in your calculations.

Spams and Plagiarism? NO.​

Answer 1

Answer :-

(i) Resistance of the coil = 2.08 Ω

(ii) Potential difference = 4.16 V

The main assumption that helped us to solve this problem is "Law of conservation of energy" i.e. the energy remains constant during transfer of the heat .

Explanation :-

We have :-

→ Heat capacity of calorimeter = 50 J/°C

→ Specific heat capacity (liquid) = 450 J/kg °C

→ Mass of the liquid = 1 kg

→ Rise in temperature = 10°C

→ Current = 2 A

→ Time = 10 min = 600 sec

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Firstly, let's calculate the amount of heat energy lost by the coil .

H = I²Rt

⇒ H = (2)² × R × 600

⇒ H = 4 × R × 600

⇒ H = 2400R ----(1)

Now, heat gained will be :-

= Heat gained (calori.) + Heat gained (liq.)

= C₁∆T + mC₂∆T

= 50 × 10 + 1 × 450 × 10

= 5000 J

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According to principle of calorimtery, the heat energy will reamain conserved .

⇒ Heat lost = heat gained

⇒ 2400R = 5000

⇒ R = 5000/2400

⇒ R = 2.08 Ω

Finally, as per Ohm's Law :-

V = IR

⇒ V = 2 × 2.08

⇒ V = 4.16 V

Answer 2

I have attached your Answer.

hope it helps you!!