Question

An electric heater of resistance 8Ω draws 15A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.​

Answer 1

ɢɪᴠᴇɴ:-

Resistane (R) = 8 Ω

Electric current (I) = 15 A

Time (T) = 2 h = 2 × 60 × 60 = 7200 sec

ᴛᴏ ꜰɪɴᴅ:-

Electric Power (P)

ꜱᴏʟᴜᴛɪᴏɴ:-

As we know that,

$\red{\boxed{\gray{\sf{H = I^2 R T}}}}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {H\: = (15)^2 \times 8 \times 7200} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {H \: = 225 \times 57,600} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {H = 12,960,000} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak\green{H \: = \: 12,960,000 \: J} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\tt \pink{\boxed{\purple{\sf{Power = \frac{Heat}{Time}}}}}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {Power = \frac{12960000}{7200}} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {Power = 1800} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak\pink{Power \: = \: 1800 \: Js^{-1}} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

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Answer 2

$\huge\boxed{\fcolorbox{purple}{ink}{Answer}}$

The formula unit mass of:

(i) ZnO = 65 u + 16 u = 81 u

(ii) Na2O = (23 u x 2) + 16 u = 46 u + 16 u = 62 u

(iii) K2C03 = (39 u x 2) + 12 u + 16 u x 3

= 78 u + 12 u + 48 u = 138 u