Question

The resistance of an electric geyser is 14 ohms. Current flowing is 4A. How much water can it heat to raise its temperature by ${10^ \circ }C$ in 10 min?
Use specific heat capacity of water = $4.186 \times {10^3}Jk{g^{ - 1}}{}^ \circ {C^{ - 1}}$
A. 0.5 kg
B. 1.6 kg
C. 3.2 kg
D. 2.4 kg

Answer 1

Hint: Before we get to the unit of specific heat, it is important that we understand the basic definition of heat capacity.
The heat capacity of a material is defined as the rate of change of heat energy in the system when there is a change in the temperature.
$C = \dfrac{{\Delta Q}}{{\Delta t}}$
where $\Delta Q$ is the change in heat energy and $\Delta t$ is the change in temperature.

Complete step-by-step answer:
Heat capacity is an extensive property and its corresponding intensive property is called specific heat capacity.
The total heat energy, $Q = ms\Delta t$
where s = specific heat capacity in $J{K^{ - 1}}k{g^{ - 1}}$ and m = mass in kg.
So, the specific heat can be defined as the amount of heat needed to raise the temperature of unit mass by a certain amount.
The electric energy, $E = {I^2}Rt$ where I is current, R is resistance and t is time.
The heat energy produced in this case is due to electric current.
Thus, $Q = E$
$ms\Delta t = {I^2}Rt$
Given:
Specific heat,$s = 4.186 \times {10^3}Jk{g^{ - 1}}{}^ \circ {C^{ - 1}}$
Temperature difference,$\Delta t = {10^ \circ }C$
Current, $I = 4A$
Resistance, $R = 14\Omega $
Time, $t = 10\min = 600\sec $
Substituting in the equation, we get –
$
  ms\Delta t = {I^2}Rt \\
  m \times 4.186 \times {10^3} \times 10 = {4^2} \times 14 \times 600 \\
  Solving, \\
  m \times 41860 = 134400 \\
   \to m = \dfrac{{134400}}{{41860}} = 3.2kg \\
 $

Therefore, the correct option is Option C.

Note: The students should be very careful about the units of specific heat capacity in the questions.
The SI unit for specific heat capacity is $J{K^{ - 1}}k{g^{ - 1}}$.
The other unit in CGS system is $cal/g/{}^ \circ C$
The rate of conversion is:
1 $cal/g/{}^ \circ C = 4187J{K^{ - 1}}k{g^{ - 1}}$