A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than

Q: Why do we assume a steady state?

The phrase 'after a long time' indicates that the system has had enough time to reach thermal equilibrium. At this point, the temperature stops rising because the heat gained from the sun exactly equals the heat lost to the surroundings.

Q: Why is the unit of the cooling constant $s^{-1}$?

In Newton's Law of Cooling, $\frac{dT}{dt} = -k \Delta T$. The left side has units of Kelvin per second ($K/s$), and $\Delta T$ has units of Kelvin ($K$). For the equation to balance, $k$ must have units of $1/s$ or $s^{-1}$.

Q: Do we need to convert the temperature difference to Kelvin?

No, a temperature difference ($\Delta T$) is exactly the same in both Celsius and Kelvin scales. An increase of $1^\circ\text{C}$ is identical to an increase of $1\text{ K}$.

Q: Why did we ignore the container's heat capacity?

The problem explicitly stated to 'Ignore effect of the container'. If we had to include it, the total heat capacity would be $(m_{\text{water}}s_{\text{water}} + m_{\text{container}}s_{\text{container}})$, but the steady-state temperature difference would actually remain the same because the heat capacity term cancels out when finding the final equilibrium state if $k$ is defined for the whole system.

Q: What if the area for heat loss was given separately?

In this specific problem, the cooling constant $k$ already inherently accounts for the surface area and emissivity of the radiating surface. If you were given the Stefan-Boltzmann constant instead, you would need the exact surface area to calculate the heat loss.

Question

A container with $1	ext{ kg}$ of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is $700	ext{ Wm}^{-2}$ and it is absorbed by the water over an effective area of $0.05	ext{ m}^2$. Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in $^\circ	ext{C}$) in the temperature of water and the surroundings after a long time will be ________. (Ignore effect of the container, and take constant for Newton's law of cooling = $0.001	ext{ s}^{-1}$, Heat capacity of water = $4200	ext{ J kg}^{-1}	ext{K}^{-1}$)

Sigma Percentile · Accepted Answer

Imagine a container filled with one kilogram of water, sitting out in the sun. It's constantly absorbing energy from the sunlight, but at the same time, it's losing heat to the cooler surroundings. This is a classic thermodynamics problem that beautifully marries calorimetry with heat transfer.

### Analyzing the Setup

The question asks for the temperature difference 'after a long time'. In the realm of physics, 'after a long time' is a massive neon sign pointing towards a **steady state**.

At steady state, the system has reached a dynamic equilibrium. The rate at which the water absorbs heat from the sun perfectly balances the rate at which it loses heat to the surroundings. Because the energy in equals the energy out, the temperature of the water stops changing and becomes constant.

### The Master Equation

Let's break down the two sides of our energy balance equation. First, we calculate the power absorbed by the water. The power absorbed is simply the intensity of the sunlight multiplied by the effective area over which it falls.

$$P_{	ext{absorbed}} = I 	imes A_{	ext{eff}}$$

Substituting the given values, we get:

$$P_{	ext{absorbed}} = 700 	imes 0.05 = 35	ext{ W}$$

So, the water is continuously receiving $35	ext{ Joules}$ of energy every second. Now, what about the heat loss? The problem states it follows **Newton's Law of Cooling**. According to this law, the rate of change of temperature is proportional to the temperature difference between the body and its surroundings.

$$\frac{dT}{dt} = -k(T - T_0) = -k\Delta T$$

But we need the rate of heat loss (which is power), not just the rate of temperature change. Remember from calorimetry that heat $Q = ms\Delta T$. Therefore, the rate of heat loss is the mass times the specific heat times the rate of cooling.

$$P_{	ext{loss}} = \left| m s \frac{dT}{dt} ight| = m s k \Delta T$$

### Final Calculation

Now, we bring it all together by equating the power absorbed to the power lost.

$$P_{	ext{absorbed}} = P_{	ext{loss}}$$

$$35 = 1 	imes 4200 	imes 0.001 	imes \Delta T$$

Look closely at the right side. $4200$ multiplied by $0.001$ is simply $4.2$. So, our equation simplifies beautifully:

$$35 = 4.2 	imes \Delta T$$

Finally, dividing $35$ by $4.2$ gives us our answer.

$$\Delta T = \frac{35}{4.2} = \frac{350}{42} = \frac{25}{3} \approx \mathbf{8.33^\circ	ext{C}}$$

The steady-state temperature difference is $8.33^\circ	ext{C}$. A perfect balance of energy!

Enter Numerical Value:

Visualized Solution

$Visualizing the System$

$The Steady State Condition$

$Calculating Power Absorbed$

$Power Absorbed Result$

$Newton’s Law of Cooling$

$Power Lost Equation$

$Equating Powers$

$Simplifying the Equation$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Newton’s Law of Cooling

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

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

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

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

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