A liquid in a beaker has temperature theta(t) at time t and theta_0 is temperature of surroundings,

Q: Why is there a negative sign in Newton's Law of Cooling?

The negative sign indicates that the temperature of the body is decreasing over time. Since the rate of change of temperature $\frac{d\theta}{dt}$ is negative for a cooling body, we add a minus sign to make the right side of the equation positive, matching the positive constant $k$.

Q: What does the constant $C$ represent in the integrated equation?

The constant $C$ is the constant of integration. Physically, it relates to the initial temperature of the liquid. At $t = 0$, $\ln(\theta_{initial} - \theta_0) = C$. So, $C$ determines the y-intercept of our straight line graph.

Q: Why do we plot the log of the temperature difference instead of just the temperature?

Plotting the raw temperature $\theta$ against time $t$ gives an exponential decay curve, which is hard to analyze visually. By plotting the logarithm of the temperature difference, we linearize the data. A straight line is much easier to verify experimentally, and its slope directly gives us the cooling constant $k$.

Q: What would the graph look like if we plotted $\theta$ vs $t$?

If you plot $\theta$ against $t$, you would get an exponentially decreasing curve that starts at the initial temperature $\theta_{initial}$ and asymptotically approaches the surrounding temperature $\theta_0$ as time goes to infinity.

Q: Does Newton's Law of Cooling hold for very large temperature differences?

Newton's Law of Cooling is actually an approximation that works best for small temperature differences between the body and its surroundings. For very large temperature differences, heat loss due to radiation (governed by Stefan-Boltzmann Law, which depends on $T^4$) becomes significant, and the simple linear approximation fails.

Question

A liquid in a beaker has temperature $	heta(t)$ at time $t$ and $	heta_0$ is temperature of surroundings, then according to Newton's law of cooling, the correct graph between $\log_e(	heta - 	heta_0)$ and $t$ is

Sigma Percentile · Accepted Answer

### The Cooling Cup of Tea

Imagine you just poured yourself a steaming cup of tea. If you leave it on the table, it doesn't stay hot forever; it gradually cools down until it reaches the temperature of the room. This everyday phenomenon is perfectly described by **Newton's Law of Cooling**.

The law states that the rate at which an object loses heat is directly proportional to the difference in temperature between the object and its surroundings. Mathematically, we write this as:

$$-\frac{d	heta}{dt} = k(	heta - 	heta_0)$$

Here, $	heta$ is the temperature of the liquid at any time $t$, $	heta_0$ is the constant temperature of the surroundings, and $k$ is a positive proportionality constant. The negative sign is crucial—it tells us that the temperature is decreasing as time goes on.

### Setting Up the Math

To find out exactly how the temperature changes over time, we need to solve this differential equation. The first step is to separate the variables. We gather all the temperature terms on one side and the time terms on the other:

$$\frac{d	heta}{	heta - 	heta_0} = -k dt$$

Now, we are ready to integrate both sides.

$$\int \frac{d	heta}{	heta - 	heta_0} = -k \int dt$$

### The Power of Logarithms

Integrating the left side gives us a natural logarithm, and the right side gives us a simple linear term in $t$:

$$\ln(	heta - 	heta_0) = -kt + C$$

where $C$ is our constant of integration.

Take a moment to look at this equation. It might look a bit abstract, but it actually has a very familiar structure. Let's compare it to the standard equation of a straight line, $y = mx + c$.

### Decoding the Graph

If we set our y-axis to be $y = \ln(	heta - 	heta_0)$ and our x-axis to be $x = t$, the equation matches perfectly!

Our slope $m$ corresponds to $-k$, and our y-intercept $c$ corresponds to $C$. Because $k$ is a positive constant, our slope $-k$ is strictly negative.

Therefore, if we plot $\log_e(	heta - 	heta_0)$ against time $t$, we will get a **straight line with a negative slope**.

This is a brilliant trick used often in experimental physics. While plotting the raw temperature $	heta$ against time $t$ yields a tricky exponential decay curve, taking the logarithm transforms the data into a straight line, making it incredibly easy to verify the law and calculate the cooling constant $k$ directly from the slope!

Animated Solution for Physics - Properties of Solids and Liquids: A liquid in a beaker has temperature $θ (t)$ at time $t$ and $θ_{0}$ is temperature of surroundings, then according to Newton's law of cooling, the correct graph between $lo g_{e} (θ - θ_{0})$ and $t$ is

Select Answer:

Visualized Solution

Newton's Law of Cooling

Differential Equation

Rearranging Variables

Integration

The Linear Equation

Equation of a Straight Line

Exponential Decay

The Sigma Insight: Heat Transfer

The Cooling Cup of Tea

Setting Up the Math

The Power of Logarithms

Decoding the Graph

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