Question

A resistance coil of $60\Omega$ is immersed in $42000\text{gm}$ of water. A current of $7\text{A}$ is passed through it. The rise in temperature of water per minute is (Specific heat of water $=4200{\text{Jkg}}^{-1}{\text{K}}^{-1}$)

• A. 1
• B. 1.0
• C. 1.00

Answer 1

Answer: (A), (B), (C)

$m=42.000\text{kg}i=7\text{A};R=60\Omega$

$H=i^{2}Rt...\left(1\right)$

$H=ms\Delta \theta ...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get,

$ms\Delta \theta =i^{2}Rt$

$\frac{\Delta \theta }{t}=\frac{i^{2}R}{ms}=\frac{7^2\times 60}{42\times 4200}$

$\frac{\Delta \theta }{t}=\frac{6}{6\times 6\times 10}=\frac{1}{60}{}^{\circ }\text{C/s}$

For $1$ minute,

$\frac{\Delta \theta }{t}=\frac{1}{60}\times 60{}^{\circ }\text{C/min}=1^{\circ }\text{C}$