Question

How many joules of heat are lost by 3580 kg of granite as it cools from 41.2°C to -12.9°C? The specific heat of granite is 0.803 J/(g·°C).

Short answer

The amount of heat lost by the 3580 kg of granite as it cools from 41.2°C to -12.9°C is approximately \(2,897,447,940\) Joules.

Step-by-step solution

Identify the given information and the formula to use

In this problem, we are given the mass (m) of granite - 3580 kg, the initial temperature (T1) - 41.2 °C, the final temperature (T2) - (-12.9) °C, and the specific heat (c) of granite - 0.803 J/(g.°C). We need to find the amount of heat lost (Q) by the granite. We will use the heat transfer formula: Q = mcΔT where ΔT represents the difference in temperature (T1 - T2).

Convert the mass of granite to grams

Since the specific heat is given in J/(g·°C), we need to convert the mass of granite from kilograms (kg) to grams (g). 1 kg = 1000 g So, for 3580 kg of granite: mass = 3580 kg * 1000 g/kg = 3,580,000 g

Find the temperature difference ΔT

To calculate ΔT, subtract the final temperature (T2) from the initial temperature (T1): ΔT = T1 - T2 = 41.2°C - (-12.9°C) = 54.1°C

Calculate the amount of heat lost Q using the formula

Now that we have the mass of granite in grams, the temperature difference, and the specific heat, we can use the heat transfer formula to find the amount of heat lost: Q = mcΔT Q = (3,580,000 g) * (0.803 J/(g·°C)) * (54.1°C) Q = 2,897,447,940 J The amount of heat lost by the granite as it cools from 41.2°C to -12.9°C is approximately 2,897,447,940 Joules.

Key concepts

Specific Heat Capacity

Specific heat capacity is a fundamental concept in the study of heat transfer in chemistry, representing the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. It's a unique value for every material, and it's crucial when quantifying the thermal energy exchange during a temperature change. Granite, for instance, has a specific heat capacity of 0.803 J/(g·°C), which means it takes 0.803 joules to raise one gram of granite by one degree Celsius.

Understanding specific heat capacity is essential because it explains why different materials heat up or cool down at different rates. It's also a cornerstone in solving problems involving heat transfer, as seen in the textbook exercise where knowing the specific heat capacity of granite was pivotal.

When improving exercises involving specific heat capacity, ensure to emphasize the units of measurement and the significance of specific heat in daily life applications, such as designing effective cooling systems or choosing materials with suitable thermal properties for construction.

Calorimetry

Calorimetry is the measurement of heat transfer from one substance to another when they are brought into physical contact. It's a technique commonly used in chemistry to determine the heat effects of chemical reactions, phase transitions, or physical changes. The principle behind calorimetry is energy conservation—energy cannot be created or destroyed, only transferred or transformed.

In a calorimetry experiment, a calorimeter is often used. It's an insulated device that helps measure the change in temperature when a substance absorbs or releases heat. The textbook problem essentially asked us to perform a simple calorimetric calculation without the actual calorimeter: determining the energy lost by a large quantity of granite as it cooled down.

To improve the understanding of this process, one might include a practical example, such as performing an experiment using a coffee cup calorimeter, a simple yet effective educational tool. This hands-on approach can illustrate the principles of calorimetry in a tangible, unforgettable manner.

Temperature Change Calculations

Temperature change calculations are fundamental to understanding heat transfer in various fields such as chemistry, physics, and engineering. The process involves determining the change in energy, or heat (Q), as a substance undergoes a change in temperature (ΔT). The formula used is directly based on the concept of specific heat capacity:\[ Q = mcΔT \]where 'm' is the mass, 'c' is the specific heat capacity, and 'ΔT' is the temperature difference. The solution to our textbook exercise involved precisely such a calculation to find the amount of heat lost by granite as it cooled.

Enhancing these exercises involves not only the mathematical process but also the practical understanding of temperature change. Students should be guided to think about real-life examples, such as the way in which the environmental temperature can affect the cooling rate of a parked car's engine or how a lake's temperature might change from day to night. Connecting these calculations to tangible experiences can make the concept more accessible and clear, ensuring students not only perform the steps but also grasp the underlying concepts intimately.