Question

Time taken by a $840W$ heater to heat one litre of water from ${10}^{o}C$ to ${40}^{o}C$ is:

• A. $50s$
• B. $100s$
• C. $150s$
• D. $200s$

Answer 1

Answer: (C)

$150s$

We know that: $Q=W/t=mC\Delta T$

Density for wate,r $d=1000\times \frac{g}{l}$

$mass=density\times volume=1000\frac{g}{l}\times 1l=1000g$

$C$ is specific heat which is $4.186\frac{J}{g^{o}C}$ and,

$\Delta T$ is temperature which is ${30}^{o}C(40-10)$

Power produced, $P=\frac{W}{t}=836W$

So putting the values:

$836t=1000\times 4.186\times 30$

$\Rightarrow t=150s$