Question

Calculate the time required to heat 20 kg of water from ${10}^{\circ }C$ to ${35}^{\circ }C$ using an immersion heater rated 1000 W. Assume that 80% of the power input is used to heat the water. Specific heat capacity of water $=4200Jk{g}^{-1}{K}^{-1}$.

Answer 1

Step 1: Given that:

Mass(m) of water = $20kg$

Initial temperature($t_1$) = ${10}^{0}C$

Final temperature($t_2$) = ${35}^{0}C$

Thus, Change in temperature($\Delta T$) = ${35}^{0}C-{10}^{0}C={25}^{0}C$

Power of the given immersion heater(P) = $1000W$

Used input power = $80\%$

Specific heat capacity of water(s) = $4200Jk{g}^{-1}{K}^{-1}$

Step 2: formula used:

The heat required to raise the temperature of a body is given by;

$Q=ms\Delta T$

Where; $s$ is the specific heat capacity and $\Delta T$ is the change in temperature of the body.

Energy is given as; $E=P\times t$

Where; P is the output power for E is the consumed energy in time t.

Step 3: Calculation of the heat energy of the water:

$Q=20kg\times 4200Jk{g}^{-1}{K}^{-1}\times {25}^{0}C$

$Q=84000\times 25$

$Q=2100000$

$Q=21\times {10}^{5}J$

Q is the heat energy used to rise the temperature of the water.

Step 4: Calculation of the time required:

Since output power = $80\%of1000W$

$P_{output}=\frac{80}{100}\times 1000$

$P_{output}=0.8\times 1000W$

Therefore if the time taken is t then ;

$21\times {10}^{5}J=0.8\times 1000W\times t$

$t=\frac{21\times {10}^5}{800}$

$t=0.02625\times {10}^5$

$t=2625sec$

$t=\frac{2625}{60}min$

$t\approx 44min$

Therefore the required time is $44min$ .