Heat Exchange Calculation: Vegetables, Fridge & Temperature Change

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Hey guys! Let's dive into a cool physics problem involving heat transfer. We're going to calculate how much heat is exchanged when you put some hot veggies into a cold fridge. This is super relevant to everyday life – think about how your food cools down in the refrigerator. We'll break down the problem step-by-step, explaining the concepts and calculations. Ready? Let's go!

Understanding the Scenario: Vegetables, Fridge and Heat Transfer

Okay, imagine you have 250 grams of freshly cooked vegetables that are a sizzling 90°C. You decide to pop them into your refrigerator, which is chilling at a cool 5°C. The key question is: How much heat is transferred between the hot vegetables and the cold environment of the fridge? This transfer will continue until they reach thermal equilibrium which means both have the same temperature. But to figure out exactly how much heat, we also need to know some specific properties of the substances involved. The specific heat capacity is a super important concept here. It tells us how much energy (in the form of heat) it takes to raise the temperature of 1 kilogram of a substance by 1 degree Celsius. Different materials have different specific heat capacities. We are given the specific heat capacities for vegetables (0.95 kcal/kg°C) and for the food in the fridge (0.65 kcal/kg°C). Also, a key concept that we must understand is that the heat lost by the vegetables must be equal to the heat gained by the fridge's internal environment (assuming no heat loss to the outside). The whole purpose of the fridge is to extract heat and maintain its internal temperature, so this makes sense. The final step is to calculate the change in temperature for both the vegetables and the food inside the refrigerator.

Now, let's break down the problem into smaller, manageable chunks so that we can clearly understand the concepts: First, we will calculate the heat lost by the vegetables, then we will calculate the heat gained by the food in the fridge. We will use the formula Q = mcΔT, where:

  • Q = heat transferred (in kcal)
  • m = mass (in kg)
  • c = specific heat capacity (in kcal/kg°C)
  • ΔT = change in temperature (Tfinal - Tinitial) in °C

So, grab a pen and paper because we're about to crunch some numbers! This exercise is a great illustration of the laws of thermodynamics at work, and it's something we encounter daily without even realizing it. The most important thing to remember is the principle of conservation of energy: the heat lost by one object (the vegetables) is equal to the heat gained by another (the food and the fridge environment). This means there's no energy being created or destroyed, just transferred. This is the foundation of understanding how heat transfer works and is crucial in many areas, from engineering to cooking! Now, let’s get into the calculation. Prepare to flex those mental muscles!

Calculating Heat Exchange: Vegetables and the Fridge Environment

Alright, let's get down to the nitty-gritty and do the calculations. We have our hot vegetables (90°C) and the cool fridge (5°C). We'll start by figuring out the heat lost by the vegetables. Remember the formula we talked about? Q = mcΔT. We know the mass of the vegetables is 250 grams, which is 0.25 kg (since 1 kg = 1000 g). The specific heat of the vegetables is given as 0.95 kcal/kg°C. We also know the initial temperature of the vegetables (90°C), and we need to figure out the final temperature they reach. In this ideal scenario, we will consider the final temperature to be very close to the temperature of the fridge, at equilibrium. Since the fridge is so much bigger, the final temperature will almost be the same temperature as the fridge. So let's approximate the final temperature to be 5°C. Now, let's plug those values into the formula to calculate the heat lost by the vegetables.

Q_vegetables = m_vegetables * c_vegetables * (T_final - T_initial)

Q_vegetables = 0.25 kg * 0.95 kcal/kg°C * (5°C - 90°C)

Q_vegetables = 0.25 kg * 0.95 kcal/kg°C * -85°C

Q_vegetables = -20.125 kcal

So, the vegetables lose 20.125 kcal of heat. This negative sign just means the heat is leaving the vegetables. Now, remember the principle of conservation of energy: the heat lost by the vegetables is gained by the food inside the fridge. So, Q_food = 20.125 kcal. The refrigerator environment will gain this heat. However, the temperature change within the refrigerator will depend on the total mass of the food inside the fridge. Since we don't have this, we can only talk about the heat absorbed. If the fridge was completely empty, all the heat would go into the walls. Therefore, to calculate the temperature change of the food, we need to know the mass of the food and its specific heat capacity (which is given as 0.65 kcal/kg°C).

Let’s now calculate the temperature change for the food. We will rearrange the formula Q = mcΔT to solve for ΔT: ΔT = Q / mc. We already know the heat gained by the food (20.125 kcal) and its specific heat (0.65 kcal/kg°C). But we need to know the mass of the food in the fridge. Let's assume there is a total of 1 kg of food in the fridge.

ΔT_food = Q_food / (m_food * c_food)

ΔT_food = 20.125 kcal / (1 kg * 0.65 kcal/kg°C)

ΔT_food = 31°C

This means that the temperature of the food inside the refrigerator would increase by 31°C, from 5°C to 36°C. However, refrigerators are designed to maintain a constant temperature, so the fridge's compressor will work to remove this heat and keep the temperature at 5°C. And this also assumes there is only 1kg of food inside the refrigerator, which is unlikely. So, in this calculation, we are assuming that the fridge's internal environment is what is affected. But the key takeaway is this: the hot vegetables transfer heat to the food inside the fridge (and the fridge environment) until they reach thermal equilibrium.

Temperature Changes and Implications: Vegetables and the Fridge

As we’ve seen, the vegetables lose heat and cool down, and the food in the fridge gains heat, potentially increasing its temperature. The size of this temperature change depends on how much food is in the fridge. In a real-world scenario, the fridge is constantly working to maintain its temperature. The food inside the fridge, and the environment of the fridge, absorb the heat from the hot vegetables. The fridge's cooling system (the compressor) then works to remove this heat, preventing a significant temperature rise. So, the impact of putting hot vegetables in the fridge is that the fridge's cooling system needs to work harder to maintain its set temperature. If you put too many hot items in at once, the fridge might struggle to keep everything cold enough, and the food could spoil faster. This also means more energy is consumed by the refrigerator, impacting your electricity bill!

The temperature of the vegetables decreases from 90°C to near 5°C and the food absorbs the heat and the temperature changes. The change in temperature (ΔT) that the food experiences depends on its mass and specific heat. In a real fridge, the temperature change is minimal due to the fridge's ability to extract heat. However, if you were to put a very large amount of hot food in a small, poorly insulated fridge, the temperature could indeed increase significantly before the fridge could bring it back down. This highlights the importance of not overloading your refrigerator with hot items. Allowing food to cool down before putting it in the fridge is always a good idea, as it minimizes the burden on the cooling system and helps the refrigerator operate more efficiently. Cool food helps maintain the consistent cold environment that is essential for preserving food quality and safety.

Summary and Key Takeaways

Alright, let's recap what we've learned! We calculated the heat exchange between hot vegetables and the cold environment of a refrigerator. We used the formula Q = mcΔT and considered the principle of conservation of energy. We also touched upon the impact on the fridge's cooling system. Here's a quick summary:

  • Heat Transfer: Hot vegetables transfer heat to the fridge's internal environment.
  • Specific Heat: The specific heat capacity tells us how much heat a substance needs to change its temperature.
  • Calculations: We calculated the heat lost by the vegetables and the potential temperature change inside the fridge.
  • Real-World Implications: Putting hot food in the fridge can make the cooling system work harder and potentially affect food preservation.

So, the next time you put leftovers in the fridge, remember this heat transfer process! It's a fundamental concept in physics that affects our everyday lives. Keep in mind that food should cool a bit before going in to help the refrigerator maintain its temperature. Hope this was useful, guys! Keep those questions coming! Until next time, stay cool and keep exploring the amazing world of science! The beauty of this is that the same principles apply whether we are talking about vegetables, a cup of coffee, or the engine of a car. These are fundamental and can be applied in various situations.