Question

An immersion heater is rated $836{\text{W}}$ . Find the time taken for it to heat $1{\text{L}}$ of water from $20^\circ {\text{C}}$ to $40^\circ {\text{C}}$ .
A. $200{\text{s}}$
B. $100{\text{s}}$
C. ${\text{836s}}$
D. ${\text{418s}}$

Answer 1

Hint: Power is defined as the energy or work done over time. Heat is a form of energy. The heat generated by the heater is supplied to the given volume of water. So the heat required to change the temperature of the water from $20^\circ {\text{C}}$ to $40^\circ {\text{C}}$ will be equal to the heat produced by the heater in the required time

Formulae Used:
->The amount of heat required to cause a temperature change $\Delta T$ in a given sample of mass is given by, $Q = ms\Delta T$ where $m$ is the mass of the sample and $s$ is the specific heat capacity of the sample.
->The power produced is given by, $P = \dfrac{Q}{t}$ where $Q$ is the heat generated in time $t$ .

Complete step-by-step solution:
Step 1: List the known parameters.
The power of the heater is given to be $P = 836{\text{W}}$ .
The required change in temperature is $\Delta T = 40 - 20 = 20^\circ {\text{C}}$ .
The volume of water to be heated is $V = 1{\text{L}} = {\text{0}} \cdot {\text{001}}{{\text{m}}^3}$ .

Step 2: Express the relation for the mass of the water to be heated.
The mass of water is given by, $m = \rho V$ ------- (1)
Substituting for $V = {\text{0}} \cdot {\text{001}}{{\text{m}}^3}$ and $\rho = 1000{\text{kg}}{{\text{m}}^{ - 3}}$ in equation (1) we get, $m = 0 \cdot 001 \times 1000 = 1{\text{kg}}$
So the mass of the given volume of water is $m = 1{\text{kg}}$ .

Step 3: Express the relation for the heat required to cause the given change in temperature.
The amount of heat required to cause the given temperature change is given by,
$Q = ms\Delta T$ --------- (2)
where $m$ is the mass and $s$ is the specific heat capacity of water.
For water the specific heat capacity is $s = 4182{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$
Now substituting for $m = 1{\text{kg}}$ , $s = 4182{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$ and $\Delta T = 20{\text{K}}$ in equation (2) we get, $Q = 1 \times 4182 \times 20 = 83640{\text{J}}$
Thus the amount of heat produced by the heater will be $Q = 83640{\text{J}}$ .

Step 4: Using the relation for the power of the heater, find the time taken to cause the mentioned change in temperature.
The power of the heater is given by, $P = \dfrac{Q}{t}$ where $Q$ is the heat generated in time $t$ .
$ \Rightarrow t = \dfrac{Q}{P}$ ------- (3)
Substituting for $Q = 83640{\text{J}}$ and $P = 836{\text{W}}$ in equation (3) we get, $t = \dfrac{{83640}}{{836}} = 100 \cdot 04{\text{s}}$
Thus the time taken to cause the mentioned change in temperature is $100{\text{s}}$ .
Hence the correct option is B.

Note:- While substituting values for physical quantities in equations(1), (2) and (3) make sure that the physical quantities are expressed in their respective S.I. units. If not, then the necessary conversion of units must be done. The temperature change is a difference and so does not have to be converted to its S.I. unit as the difference in temperature will be the same in the Kelvin scale.