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\(\cdot\) You have 750 \(\mathrm{g}\) of water at \(10.0^{\circ} \mathrm{C}\) in a large insulated beaker. How much boiling water at \(100.0^{\circ} \mathrm{C}\) must you add to this beaker so that the final temperature of the mixture will be \(75^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
Add 1950 grams of boiling water.

Step by step solution

01

Understand the Concept

In this problem, we are dealing with heat exchange. The idea is that energy lost by the hot water while cooling down to the final temperature will be equal to the energy gained by the cold water as it heats up to the final temperature. We will use the principle of conservation of energy.
02

Identify Key Equations

The key equation for heat gained or lost is \[ q = mc\Delta T \]where \( q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat capacity (for water it's 4.18 J/g°C), and \( \Delta T \) is the change in temperature.
03

Set Up the Equation

Let \( m_h \) be the mass of the hot water to be added. The heat lost by the hot water is:\[ q_h = m_h \times c \times (100.0^{\circ} - 75.0^{\circ}) \]The heat gained by the cold water is:\[ q_c = 750 \times c \times (75.0^{\circ} - 10.0^{\circ}) \]Set \( q_h = q_c \) for conservation of energy.
04

Simplify and Solve the Equation

Equating the two expressions gives:\[ m_h \times c \times 25 = 750 \times c \times 65 \]The specific heat \( c \) cancels out:\[ m_h \times 25 = 750 \times 65 \]Thus, solve for \( m_h \):\[ m_h = \frac{750 \times 65}{25} \]
05

Calculate the Mass of Boiling Water Needed

Perform the calculation:\[ m_h = \frac{48750}{25} = 1950 \]Therefore, 1950 grams of boiling water must be added to achieve the desired final temperature.

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is a fundamental concept in physics that tells us that energy cannot be created or destroyed. Instead, it can only change forms. In the context of our problem, this means that the heat energy lost by the hot water is equal to the heat energy gained by the cold water. This is crucial to find out how much boiling water we need to add.

By setting the heat lost by the hot water equal to the heat gained by the cold water, we can solve for the unknown mass of the hot water. This ensures that energy is balanced in the closed system of the insulated beaker, where no heat is lost to the surroundings.
Specific Heat Capacity
Specific heat capacity is a property of a substance that tells us how much heat energy is needed to raise the temperature of one gram of the substance by one degree Celsius.

For water, this value is 4.18 J/g°C. This means it takes 4.18 joules of energy to raise one gram of water by one degree Celsius. Knowing this value allows us to calculate the amount of heat energy exchanged during the process of heating or cooling.

It’s essential to understand this concept because it explains how different materials respond to heat. In our exercise, using the specific heat capacity of water, we plug it into the heat equation to calculate the energy changes when mixing different temperatures of water.
Temperature Change
Temperature change is represented as \( \Delta T \) in the heat equation \( q = mc\Delta T \). It indicates how much the temperature of a substance has increased or decreased.

In our exercise, the hot boiling water undergoes a decrease in temperature from \( 100^{\circ} \text{C} \) to the final temperature \( 75^{\circ} \text{C} \). Meanwhile, the cold water's temperature increases from \( 10^{\circ} \text{C} \) to \( 75^{\circ} \text{C} \).
  • The temperature change for the hot water: \( \Delta T = 100 - 75 \)
  • For the cold water: \( \Delta T = 75 - 10 \)
Understanding this helps us apply changes in temperature to determine the heat energy transferred in each scenario. It’s a key step in the process of calculating the mass of boiling water needed.
Mass Calculation
Mass calculation can be easily done when we understand all the above concepts and have our equations correctly set up. In our exercise, we calculated the mass of the boiling water needed to reach a desired final temperature with the equation:
  • \[ m_h \times 25 = 750 \times 65 \]
  • Solving gives \( m_h = \frac{750 \times 65}{25} \)
  • Therefore, \( m_h = 1950 \) grams
This calculation was straightforward because the specific heat capacity \( c \) was the same for both masses of water and cancels out, leaving a simple multiplication equation to solve for \( m_h \). Understanding this step shows how knowing a few key details about the properties of substances allows us to deduce necessary quantities in heat exchange processes.

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Most popular questions from this chapter

\(\bullet\) The effect of urbanization on plant growth. A study published in July 2004 indicated that temperature increases in urban areas in the eastern United States are causing plants to bud up to 7 days early compared with plants in rural areas just a few miles away, thereby disrupting biological cycles. Average temperatures in the urban areas were up to 3.5 \(\mathrm{C}^{\circ}\) higher than in the rural areas. By what percent will the radiated heat per square meter increase due to such a temperature difference if the rural temperature was \(0^{\circ} \mathrm{C}\) the average?

\bullet A 15.0 g bullet traveling horizontally at 865 \(\mathrm{m} / \mathrm{s}\) passes through a tank containing 13.5 \(\mathrm{kg}\) of water and emerges with a speed of 534 \(\mathrm{m} / \mathrm{s}\) . What is the maximum temperature increase that the water could have as a result of this event?

Shivering. You have no doubt noticed that you usually shiver when you get out of the shower. Shivering is the body's way of generating heat to restore its internal temperature to the normal \(37^{\circ} \mathrm{C},\) and it produces approximately 290 \(\mathrm{W}\) of heat power per square meter of body area. \(\mathrm{A} 68 \mathrm{kg}(150 \mathrm{lb}), 1.78 \mathrm{m}\) (5 foot, 10 inch) person has approximately 1.8 \(\mathrm{m}^{2}\) of surface area. How long would this person have to shiver to raise his or her body temperature by \(1.0 \mathrm{C}^{\circ},\) assuming that none of this heat is lost by the body? The specific heat capacity of the body is about 3500 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K})\) .

\(\cdot\) In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a 200.0 \(\mathrm{W}\) electric immersion heater in 0.320 \(\mathrm{kg}\) of water. (a) How much heat must be added to the water to raise its temperature from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C} ?\) (b) How much time is required if all of the heater's power goes into heating the water?

A technician measures the specific heat capacity of an unidentified liquid by immersing an electrical resistor in it. Electrical energy is converted to heat, which is then transferred to the liquid for 120 s at a constant rate of 65.0 W. The mass of the liquid is \(0.780 \mathrm{kg},\) and its temperature increases from \(18.55^{\circ} \mathrm{C}\) to \(22.54^{\circ} \mathrm{C}\) . (a) Find the average specific heat capacity of the liquid in this temperature range. Assume that negligible heat is transferred to the container that holds the liquid and that no heat is lost to the surroundings. (b) Suppose that in this experiment heat transfer from the liquid to the container or its surroundings cannot be ignored. Is the result calculated in part (a) an overestimate or an underestimate of the average specific heat capacity? Explain.

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