A water cooler of storage capacity 120 litres can cool water at a constant rate of P watts. In a clo

Q: Why do we subtract the heat absorbed by the water?

The water acts as a thermal buffer. It absorbs some of the heat generated by the device, meaning the cooler doesn't have to remove 100% of the generated heat immediately. It only needs to remove the excess.

Q: How did we get the mass of the water as 120 kg?

The density of water is given as $1000 \text{ kg/m}^3$. Since 1000 liters make $1 \text{ m}^3$, the density is exactly $1 \text{ kg/liter}$. Therefore, 120 liters of water has a mass of 120 kg.

Q: Why do we convert 3 hours into seconds?

Power is given in Watts, which is Joules per second. To keep our units consistent (Joules, kg, seconds), we must convert the time from hours to seconds by multiplying by 3600.

Q: What if the water was already at 30°C initially?

If the water was already at its maximum allowed temperature, it couldn't absorb any more heat ($\Delta T = 0$). The cooler would then have to remove all the heat generated by the device, requiring a power of exactly 3000 W.

Q: Why is the specific heat taken as 4200 instead of 4.2?

The specific heat is given as $4.2 \text{ kJ kg}^{-1}\text{K}^{-1}$. The 'k' stands for kilo ($10^3$). To convert it to standard SI units (Joules), we multiply by 1000, giving $4200 \text{ J kg}^{-1}\text{K}^{-1}$.

Question

A water cooler of storage capacity 120 litres can cool water at a constant rate of $P$ watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed 30°C and the entire stored 120 litres of water is initially cooled to 10°C. The entire system is thermally insulated. The minimum value of $P$ (in watts) for which the device can be operated for 3 hours is
(Specific heat of water is $4.2 	ext{ kJ kg}^{-1}	ext{K}^{-1}$ and the density of water is $1000 	ext{ kg m}^{-3}$)

Options:

(A) 1600

(B) 2067

(C) 2533

(D) 3933

Sigma Percentile · Accepted Answer

Imagine you are tasked with keeping a high-power 3 kW device from overheating. It's generating heat relentlessly, like a small furnace. To keep it safe, you set up a closed-loop water cooling system. But here is the catch: you don't need a cooler that can handle the full 3 kW. Why? Because you have a secret weapon—water.

### The Master Equation: Conservation of Energy

The core of this problem lies in the beautiful principle of energy conservation. Over the span of 3 hours, the device will generate a massive amount of heat. Where does this heat go? It has only two destinations:
1. It raises the temperature of the circulating water.
2. It is actively extracted from the system by the cooler.

We can write this mathematically as:
$$Q_{	ext{gen}} = Q_{	ext{water}} + Q_{	ext{cooler}}$$

Let's calculate the total heat generated first. The device operates at $3 	ext{ kW}$ (or $3000 	ext{ W}$) for 3 hours. Since 1 Watt is 1 Joule per second, we must convert the time into seconds:
$$Q_{	ext{gen}} = 3000 	imes (3 	imes 3600) = 32.4 	imes 10^6 	ext{ J} = 324 	imes 10^5 	ext{ J}$$

### The Thermal Buffer: Water's Superpower

Water has one of the highest specific heat capacities of any common substance. This makes it an incredible 'thermal buffer'. We have 120 liters of water, which translates to a mass of $120 	ext{ kg}$ (since the density of water is $1000 	ext{ kg/m}^3$).

The water enters the system at $10^\circ	ext{C}$ and is allowed to heat up to $30^\circ	ext{C}$. The heat it can absorb is given by the calorimetry formula:
$$Q_{	ext{water}} = m s \Delta T$$

$$Q_{	ext{water}} = 120 	imes 4200 	imes (30 - 10)$$

$$Q_{	ext{water}} = 10.08 	imes 10^6 	ext{ J} = 100.8 	imes 10^5 	ext{ J}$$

Notice how much heat the water simply 'soaks up' without any active cooling!

### The Cooler's Burden

Now, we find out how much work is left for the cooler. We subtract the heat absorbed by the water from the total heat generated:
$$Q_{	ext{cooler}} = Q_{	ext{gen}} - Q_{	ext{water}}$$

$$Q_{	ext{cooler}} = 324 	imes 10^5 - 100.8 	imes 10^5 = 223.2 	imes 10^5 	ext{ J}$$

This is the total energy the cooler must remove over the 3-hour period. To find the minimum power rating $P$ of the cooler, we divide this energy by the total time in seconds:
$$P = \frac{Q_{	ext{cooler}}}{t} = \frac{223.2 	imes 10^5}{10800}$$

$$P \approx 2067 	ext{ W}$$

### The Takeaway

By using 120 liters of water as a thermal buffer, we reduced the required cooling power from 3000 W down to 2067 W. This is a classic engineering optimization, beautifully illustrated through the laws of thermodynamics. The next time you see a liquid-cooled PC or a car radiator, you'll know exactly the physics keeping it from melting down!

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Visualized Solution

$System Overview$

$Principle of Calorimetry$

$Formulas$

$Heat Generated (Q_{gen})$

$Heat Absorbed by Water (Q_{water})$

$Heat Removed by Cooler (Q_{cooler})$

$Cooler Power (P)$

$Final Answer$

$The Thermal Buffer$

The Sigma Insight: Calorimetry

The Master Equation

The Thermal Buffer

The Cooler's Burden

The Takeaway

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Overview

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Principle of Calorimetry

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Formulas

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Generated (Qgen​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Absorbed by Water (Qwater​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Removed by Cooler (Qcooler​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Cooler Power (P)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Thermal Buffer