**3.**

What magnitude of current should flow through a resistor of $200 \Omega$ for 42 minutes to produce the heat necessary to raise the temperature of $7.5 ext{ kg}$ of water from $0^\circ ext{C}$ to the boiling point? The specific heat of water is $4200 ext{ J/kgK}$.

Question

**3.**

What magnitude of current should flow through a resistor of $200 \Omega$ for 42 minutes to produce the heat necessary to raise the temperature of $7.5 	ext{ kg}$ of water from $0^\circ	ext{C}$ to the boiling point? The specific heat of water is $4200 	ext{ J/kgK}$.

Accepted Answer

**AI-তৈরি নমুনা উত্তর**

**Given Data:**
- Resistance, $R = 200\ \Omega$
- Time, $t = 42	ext{ min} = 42 	imes 60 = 2520	ext{ s}$
- Mass of water, $m = 7.5	ext{ kg}$
- Initial temperature, $	heta_1 = 0^\circ	ext{C}$
- Final temperature (boiling point), $	heta_2 = 100^\circ	ext{C}$
- Temperature change, $\Delta	heta = 100 - 0 = 100^\circ	ext{C} = 100	ext{ K}$
- Specific heat of water, $s = 4200	ext{ J/(kg}\cdot	ext{K)}$

**Heat required to raise the temperature of water:**
$$Q = m s \Delta	heta$$
$$Q = 7.5 	imes 4200 	imes 100 = 3{,}150{,}000	ext{ J}$$

**According to Joule's law of heating:**
Heat energy produced in the resistor,
$$H = I^2 R t$$

Assuming all the electrical energy is converted into heat absorbed by the water ($H = Q$):
$$I^2 R t = Q$$
$$I^2 = \frac{Q}{R t} = \frac{3{,}150{,}000}{200 	imes 2520} = \frac{3{,}150{,}000}{504{,}000} = 6.25$$
$$I = \sqrt{6.25} = 2.5	ext{ A}$$

**Answer:**
The magnitude of current that should flow through the resistor is **$2.5	ext{ A}$**.

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