Question

An object is placed in a room where the temperature is \(20^{\circ} \mathrm{C}\). The temperature of the object drops by \(5^{\circ} \mathrm{C}\) in 4 minutes and by \(7^{\circ} \mathrm{C}\) in 8 minutes. What was the temperature of the object when it was initially placed in the room?

Short answer

Answer: The estimated initial temperature of the object when it was placed in the room is approximately 25.5°C.

Step-by-step solution

Determine the rate of temperature drop per minute

To find the rate at which the temperature drops per minute, we need to compare the temperature drop against time. We are given that the object's temperature drops by \(5^{\circ}\mathrm{C}\) in 4 minutes and by \(7^{\circ}\mathrm{C}\) in 8 minutes. Let's create a ratio for each scenario to find the rate of temperature drop per minute: For the 4-minute scenario: \(\frac{\Delta T}{\Delta t} = \frac{-5^{\circ}\mathrm{C}}{4 \,\text{min}}= -\frac{5}{4}^{\circ}\mathrm{C}\text{ per minute}\) For the 8-minute scenario: \(\frac{\Delta T}{\Delta t} = \frac{-7^{\circ}\mathrm{C}}{8 \,\text{min}} = -\frac{7}{8}^{\circ}\mathrm{C}\text{ per minute}\)

Calculate the initial temperature

Now that we have found the rate of temperature drop, we can calculate the initial temperature of the object. Let \(T_{init}\) represent the initial temperature of the object. Since the formula for the final temperature after a specific time period is \(T_{final} = T_{init} + \Delta T\), we can use the rate of temperature drop and room temperature to find the initial temperature. Using the 4-minute scenario, \(T_{final}\) after 4 minutes would be \(20^{\circ}\mathrm{C} + 5^{\circ}\mathrm{C} = 25^{\circ}\mathrm{C}\). Similarly, using the 8-minute scenario, \(T_{final}\) after 8 minutes would be \(20^{\circ}\mathrm{C} + 7^{\circ}\mathrm{C} = 27^{\circ}\mathrm{C}\). It's important to note that the temperature drop rate is not constant. In the 4-minute scenario, the temperature drop rate is \(-\frac{5}{4}^{\circ}\mathrm{C}\text{ per minute}\), while in the 8-minute scenario, the temperature drop rate is \(-\frac{7}{8}^{\circ}\mathrm{C}\text{ per minute}\). This means we cannot make a direct calculation using one of these rates to find the initial temperature. However, we can make an estimation. Let's take the average of the two rates. The average rate of temperature drop per minute will be: \(-\frac{5}{4}*\frac{1}{2}+\left(-\frac{7}{8}\right)*\frac{1}{2}=\frac{-5}{4}*\frac{1}{2}+\frac{-7}{8}*\frac{1}{2}=-\frac{11}{8}^{\circ}\mathrm{C}\text{ per minute}\) Now, we can calculate \(T_{init}\) by using the room temperature \(T_{final}=20^{\circ}\mathrm{C}\) and the average rate of temperature drop. Since \(T_{final} = T_{init} + \Delta T = T_{init} + \left(-\frac{11}{8}\right)*t\), we need to choose a time period, t, to calculate the initial temperature. Let's use t = 4 minutes: \(20^{\circ}\mathrm{C} = T_{init} + \left(-\frac{11}{8}\right)*4\) \(20^{\circ}\mathrm{C} = T_{init} - \frac{11}{2}\) \(T_{init} = 20^{\circ}\mathrm{C} + \frac{11}{2}=20^{\circ}\mathrm{C} + 5.5^{\circ}\mathrm{C}\) \(T_{init} = 25.5^{\circ}\mathrm{C}\) Using the average temperature drop rate, the estimated initial temperature of the object when it was placed in the room is approximately \(25.5^{\circ}\mathrm{C}\).

Key concepts

Temperature Change

The concept of temperature change is crucial when discussing thermodynamics, especially in scenarios like Newton's Law of Cooling. In the given problem, we observe a change in the temperature of an object over time within a different environment. The initial step involves identifying how much the temperature of the object reduces over specific time intervals:

For example:

• A temperature drop by \(5^{\circ} \mathrm{C}\) over 4 minutes.
• A temperature drop by \(7^{\circ} \mathrm{C}\) over 8 minutes.

This change illustrates how heat is transferred from the object to the surrounding area. The time-dependent rate at which this occurs is foundational in understanding thermal dynamics.

Temperature change helps us understand varying rates of heat loss and gain, which are influenced by factors such as the difference in temperature between the object and the environment, the surface area, and material properties of the object.

Differential Equations

Differential equations are mathematical tools used to model and describe how variables change indirectly. In the realm of Newton's Law of Cooling, they help predict, calculate, and understand how temperature evolves over time. The basic idea is to establish the relationship between the rate of temperature change concerning the object's temperature and the ambient temperature.

In the problem, the rate at which the temperature decreases was calculated per minute in the given scenarios:

• The 4-minute rate of temperature drop is \(-\frac{5}{4}^{\circ} \mathrm{C}\text{ per minute}\).
• The 8-minute rate of change is \(-\frac{7}{8}^{\circ} \mathrm{C}\text{ per minute}\).

Differential equations allow us to derive a function for temperature change over time, which can then be used to forecast temperatures at any given moment, assuming specific conditions. While the problem above does not solve a differential equation directly, understanding these rate changes is critical to forming the differential equation models in scientific applications.

Initial Temperature Estimation

Estimating the initial temperature of an object involves understanding how its temperature changes over time, especially when placed in a different environment. We apply these concepts to deduce the starting temperature before any changes occurred.

In our case, two different scenarios of temperature drop are presented. However, the rate of temperature drop is not constant.

To estimate thoroughly:

• First, calculate the average rate from provided scenarios.
• The average rate was found to be \(-\frac{11}{8}^{\circ}\mathrm{C} \text{ per minute}\).
• Using the ambient temperature and rate, calculate back to find what the initial temperature could have been using a chosen time frame.

Estimation often relies on approximations of measured rates, as exact values can vary. It's essential to use a mathematical model based on observed data and logical assumptions. The given solution used an average rate to arrive at an estimated initial temperature, which reflects common practices in thermal analysis.