Animated Solution for Physics - Properties of Solids and Liquids: Time taken by a $836\text{ W}$ heater to heat $1\text{ L}$ of water from $10^{\circ }\text{C}$ to $40^{\circ }\text{C}$ is

This problem is a beautiful intersection of electrical power and thermodynamics. It asks a very practical question: If you drop an electrical heater into a beaker of water, how long will it take to heat up? Let's break down the physics behind this everyday phenomenon.

Imagine you have a beaker containing exactly $1\text{ L}$ of water. We know from basic physics that the density of water is $1\text{ kg/L}$, which means the mass of the water is exactly $m=1\text{ kg}$.

We are dropping a heater into this water with a power rating of $P=836\text{ W}$. Power is simply the rate at which energy is supplied. This means our heater pumps out $836\text{ Joules}$ of electrical energy every single second.

Our goal is to raise the temperature of this water from an initial temperature of $T_1=10^{\circ }\text{C}$ to a final temperature of $T_2=40^{\circ }\text{C}$. This gives us a required temperature change of $\Delta T=30^{\circ }\text{C}$.

Calorimetry

To solve this, we rely on the principle of conservation of energy. Assuming the beaker is perfectly insulated and no heat escapes into the surrounding air, all the electrical energy supplied by the heater will be absorbed by the water as heat energy.

Mathematically, we can write this as:

The electrical energy supplied over a time $t$ is simply Power multiplied by time ($P\times t$). The heat energy required to raise the temperature of a substance is given by the calorimetry formula $m\times s\times \Delta T$, where $s$ is the specific heat capacity of the substance.

Equating the two, we get our master equation:

Now, we just need to plug in our known values. The specific heat capacity of water is approximately $4184{\text{ J/kg}}^{\circ }\text{C}$. However, to make calculations elegant and yield an exact integer, examiners often design problems around the value $s=4180{\text{ J/kg}}^{\circ }\text{C}$ (or they use $4200{\text{ J/kg}}^{\circ }\text{C}$ for rough estimates). Let's use $4180{\text{ J/kg}}^{\circ }\text{C}$.

First, let's calculate the total heat energy the water demands:

This means the water needs exactly $125,400\text{ Joules}$ of energy to reach $40^{\circ }\text{C}$.

Now, we equate this required heat to the energy our heater can provide:

To find the time $t$, we simply divide the total energy by the power of the heater:

It will take exactly $150\text{ seconds}$, or two and a half minutes, to heat the water.

A Quick Note on Efficiency: In the real world, no system is perfectly insulated. If the problem stated that the heater was only $80\%$ efficient, you would modify the left side of the equation to $0.8\times P\times t$. Always keep an eye out for efficiency traps in competitive exams!