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Chapter 18: Problem 4

Which will warm a childs inflatable pool more adding 500 \(\mathrm{g}\) of water at \(50^{\circ} \mathrm{C}\) or 100 \(\mathrm{g}\) at \(95^{\circ} \mathrm{C}\) ? The temperature of the water in the pool is \(20^{\circ} \mathrm{C}\) . Explain your reasoning.

Short Answer

Expert verified
500 g of water at 50掳C will warm the pool more than 100 g at 95掳C.

Step by step solution

01

Understand the Problem

We need to determine which addition of hot water - 500 g at 50掳C or 100 g at 95掳C - will increase the temperature of the pool water more. The pool's initial temperature is 20掳C.
02

Apply the Heat Transfer Formula

To compare the effects, calculate the amount of heat transferred to the pool by both quantities of additional water using the formula for heat transfer: \[ Q = mc\Delta T \]where \( m \) is mass, \( c \) is specific heat capacity (which is approximately 4.18 J/g掳C for water), and \( \Delta T \) is the temperature change of the added water.
03

Calculate Heat for 500 g of Water

Calculate the heat for 500 g of water at 50掳C:\[\Delta T = 50掳C - 20掳C = 30掳C \Q_1 = 500 \, \text{g} \times 4.18 \, \text{J/g掳C} \times 30掳C = 62,700 \, \text{J}\]
04

Calculate Heat for 100 g of Water

Calculate the heat for 100 g of water at 95掳C:\[\Delta T = 95掳C - 20掳C = 75掳C \Q_2 = 100 \, \text{g} \times 4.18 \, \text{J/g掳C} \times 75掳C = 31,350 \, \text{J}\]
05

Compare the Heat Transfers

Compare the heat added to the pool by the two water quantities. We find that 500 g of water at 50掳C transfers 62,700 J, while 100 g of water at 95掳C transfers 31,350 J. Therefore, the 500 g of water contributes more heat.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat Capacity
Specific heat capacity is a measure of how much heat energy a substance can store. It tells us how much heat is needed to increase the temperature of 1 gram of the substance by 1掳C. For water, the specific heat capacity is approximately 4.18 joules per gram per degree Celsius (J/g掳C).
This means that water can hold and transfer a large amount of heat, which is why it is often used in heat transfer applications.
  • High specific heat capacity: Water requires more energy to change temperature compared to metals.
  • Energy storage: A high specific heat means water is good at storing thermal energy, making it effective for maintaining stable temperatures.
Understanding this concept helps explain why adding more water at a lower temperature can transfer more heat than less water at a higher temperature.
Temperature Change
Temperature change is a key factor in calculating heat transfer.It is expressed as the difference between the final and initial temperatures of the substance being heated or cooled.
For the heat transfer formula \( Q = mc\Delta T \), the symbol \( \Delta T \) represents temperature change, calculated as \( \text{final temperature} - \text{initial temperature} \).
  • A larger temperature difference means more energy change.
  • In our pool example, 500 g of water at 50掳C, changing by 30掳C adds more heat than 100 g at 95掳C, changing by 75掳C.
  • The temperature change directly impacts how much energy is transferred, alongside mass and specific heat.
This understanding allows us to predict the effect of different amounts of heated water on the pool's temperature.
Mass and Energy
The mass of a substance plays a crucial role in heat transfer.When you have more mass at the same temperature change and specific heat capacity, more heat is transferred overall.
In our scenario:
  • 500 g of water provides more total heat energy than 100 g because of its larger mass.
  • The heat transfer formula highlights mass as a key factor: \( m \) (mass) directly multiplies with specific heat \( c \) and temperature change \( \Delta T \) to give energy.
  • Thus, a larger mass at a moderate temperature can release more thermal energy than a smaller mass at a higher temperature.
Understanding the role of mass can aid in anticipating which quantity of water will heat the pool more effectively.
Calorimetry
Calorimetry is the science of measuring the amount of heat transferred in chemical and physical processes.It often uses a calorimeter, although in simple scenarios like our pool, it's about comparing calculated heat energies.
With the formula \( Q = mc\Delta T \), calorimetry ensures we calculate heat changes accurately:
  • By measuring temperature changes and accounting for mass and specific heat capacity, one can figure out how much heat has moved between substances.
  • In heated pool scenarios, these calculations determine the effectiveness of added water in raising pool temperatures.
  • Calorimetry principles help assess energy efficiency and thermal properties of materials and processes.
Implementing calorimetric calculations allows us to make informed predictions on heat transfer results.

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Most popular questions from this chapter

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