NEET-XII-Physics
33: Thermal and Chemical Effects of Electric Current
- #9An immersion heater rated 1000 W, 220 V is used to heat 0.01 m3 of water. Assuming that the power is supplied at 220 V and 60% of the power supplied is used to heat the water, how long will it take to increase the temperature of the water from 15°C to 40°C?Ans : Given the operating voltage V and power consumed P, the resistance of the immersion heater,
`` R=\frac{{V}^{2}}{P}=\frac{{\left(220\right)}^{2}}{1000}=48.4\,\mathrm{\,\Omega \,}``
Mass of water, m = `` \frac{1}{100}`` × 1000 = 10 Kg
Specific heat of water, s = 4200 Jkg`` -``1 K`` -1``
Rise in temperature, θ = 25°C
Heat required to raise the temperature of the given mass of water,
Q = msθ = 10 × 4200 × 25 = 1050000 J
Let t be the time taken to increase the temperature of water. The heat liberated is only 60%. So,
`` \left(\frac{{V}^{2}}{R}\right)`` × t × 60% = 1050000 J
`` \Rightarrow \frac{(220{)}^{2}}{48.4}\times t\times \frac{60}{100}=1050000``
⇒ t = 29.17 minutes
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