Newton's Law of Cooling states that the rate of cooling of a body is directly proportional to the temperature difference between the body and its surroundings. For a liquid in a beaker with temperature θ(t) at time t and θ₀ as the surrounding temperature, the law is mathematically expressed as:

$\frac{d\theta }{dt}=-k(\theta -{\theta }_0)$

where k is a positive constant. To find the relationship between ln(θ - θ₀) and time t, we can solve this differential equation.

Step 1: Separate the variables

$\frac{d\theta }{\theta -{\theta }_0}=-kdt$

Step 2: Integrate both sides

$\int \frac{d\theta }{\theta -{\theta }_0}=-k\int dt$

This gives:

$\ln |\theta -{\theta }_0|=-kt+C$

where C is the constant of integration. Since θ > θ₀ during cooling, we can write:

$\ln (\theta -{\theta }_0)=-kt+C$

Step 3: Interpret the equation

This is of the form y = mx + c, where y = ln(θ - θ₀), m = -k, x = t, and c = C. This represents a straight line with a negative slope (-k) and an intercept C on the ln(θ - θ₀) axis.

Therefore, the graph between ln(θ - θ₀) and time t is a straight line with a negative slope.

Final Answer: The correct graph is a straight line sloping downward, which corresponds to the second option (the image showing a decreasing linear plot).

Related Topics

Newton's Law of Cooling: It describes how the temperature of an object changes over time due to heat loss to its surroundings. It is applicable when the temperature difference is small, and the cooling rate is proportional to that difference.

Formulae

The integrated form of Newton's Law of Cooling: $\ln (\theta -{\theta }_0)=-kt+C$

This can also be written as: $\theta -{\theta }_0=e^{C-kt}=A{e}^{-kt}$, where A is a constant.