Animated Solution for Physics - Properties of Solids and Liquids: Water of volume 2 L in a container is heated with a coil of 1 kW at 27°C. The lid of the container is open and energy dissipates at a rate of 160 J/s. In how much time temperature will rise from 27°C to 77°C? [Specific heat of water is 4.2 kJ/kg]

Imagine you are in a laboratory, and you have a container filled with exactly $2\text{ L}$ of water. We want to heat this water up, so we drop a powerful $1\text{ kW}$ heating coil right into it. Now, before we do any math, let's think about the mass of this water. Since the density of water is $1\text{ kg/L}$, $2\text{ L}$ means we have exactly $2\text{ kg}$ of water. Visualizing the physical setup is always the first step to solving any physics problem.

But wait, there is a catch here. The lid of our container is completely open! This means that as the coil pumps heat into the water, the water is simultaneously losing heat to the surrounding air. The problem tells us that this energy dissipates at a rate of $160\text{ J/s}$, which is the same as $160\text{ W}$. This is a classic real-world scenario where systems are never perfectly insulated.

So, what is the effective rate at which the water is actually gaining energy? It is simply the power supplied by the coil minus the power lost to the air. Let's calculate this net power. $1\text{ kW}$ is $1000\text{ W}$. Subtracting the $160\text{ W}$ lost, we get a net power of $840\text{ W}$. This $840\text{ J/s}$ is the useful energy that actually goes into heating the water.

Now, let's look at the temperature change we want to achieve. The water starts at an initial temperature of $27^{\circ }\text{C}$. We want to raise it to a final temperature of $77^{\circ }\text{C}$. If we subtract the initial from the final, we find that the change in temperature, or $\Delta T$, is exactly $50^{\circ }\text{C}$.

How much total heat energy is required to make this $50^{\circ }\text{C}$ jump? The formula for total heat is mass times specific heat times the change in temperature:

We already know the mass is $2\text{ kg}$ and the temperature change is $50^{\circ }\text{C}$. The specific heat of water is given as $4.2{\text{ kJ/kg}}^{\circ }\text{C}$. To keep our units consistent, we must convert this to $4200{\text{ J/kg}}^{\circ }\text{C}$.

Let's substitute the values and get the answer for the total heat. We multiply $2\text{ kg}$ by $4200$ by $50$. A quick mental math trick: $2\times 50$ is $100$. Then, $100\times 4200$ gives us $420,000\text{ J}$. This massive number is the total amount of energy the water needs to absorb to reach $77^{\circ }\text{C}$.

We are almost there! We know the total energy needed is $420,000\text{ J}$, and we know the net energy supplied per second is $840\text{ J/s}$. Time is simply the total heat divided by the net power:

Finally, let's look at our options. They are in minutes and seconds. So, we need to convert $500\text{ s}$ into minutes. Since one minute has $60\text{ s}$, we divide $500$ by $60$. This gives us $8\text{ min}$, which is $480\text{ s}$, leaving a remainder of $20\text{ s}$. So the final answer is $8\text{ min }20\text{ s}$. And that matches option (a) perfectly!