Question

An electric heater of power 15 KW is used 2 1/2 hours find the energy supplied by the heater in SI unit

Answer 1

Answer:-

$\red{\bigstar}$ Energy supplied $\large\leadsto\boxed{\tt\purple{1.35 \times 10^{8} \: J}}$

⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Given:-

• Power of an electric heater = 15 kW = 15 × 10³ W

• Time period for which heater is used = 2½ hr. = 150 min. = 9000 sec.

⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• To Find:-

• Energy supplied

⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Solution:-

We know,

⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\green{P = \dfrac{E}{T}}}}$

where,

• P = Power of heater

• E = Energy supplied

• T = Time period

⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Substituting in the Formula:-

➪ $\sf 15 \times 10^{3} = \dfrac{E}{9000}$

➪ $\sf E = 15 \times 10^{3} \times 9000$

➪ $\sf E = 15 \times 10^{3} \times 9 \times 10^{3}$

➪ $\sf E = 135 \times 10^{6}$

★ $\bf\pink{E = 1.35 \times 10^{8} \: J}$

⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Note:-

Joule [J] is the SI unit of heat.

⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Therefore, the energy supplied by the heater is 1.35 × 10^8 J.