/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 You have 750 g of water at 10.0\... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
You need 1950 g of boiling water to reach 75°C.

Step by step solution

01

Identify the known quantities

You know the masses and initial temperatures of both the cold and hot water. The cold water mass is 750 g at 10.0°C, and the hot water is at 100.0°C. The final temperature you want is 75°C.
02

Use the principle of conservation of energy

The heat lost by the hot water will equal the heat gained by the cold water. Express this condition as: \( m_\text{hot} \cdot c \cdot (T_\text{initial, hot} - T_\text{final}) = m_\text{cold} \cdot c \cdot (T_\text{final} - T_\text{initial, cold}) \), where \( c \) is the specific heat capacity of water.
03

Substitute known values into the equation

The specific heat capacity cancels out. Substitute the masses and temperatures: \( m_\text{hot} \cdot (100 - 75) = 750 \cdot (75 - 10) \).
04

Solve for the unknown mass of the hot water

The equation becomes \( m_\text{hot} \cdot 25 = 750 \cdot 65 \). Calculate the product on the right side and then solve for \( m_\text{hot} \): \( m_\text{hot} = \frac{750 \times 65}{25} \).
05

Calculate the final answer

Perform the calculation \( m_\text{hot} = \frac{48750}{25} = 1950 \) g. Therefore, 1950 grams of boiling water is needed.

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Key Concepts

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

Conservation of Energy
In this exercise, we see an essential principle known as the conservation of energy in action. This principle states that energy cannot be created or destroyed but only transferred or converted from one form to another. In the context of heat transfer, this means the heat lost by one substance must equal the heat gained by another.
When you mix hot and cold water, the hot water loses thermal energy or heat, which is gained by the cold water. This process continues until both reach the same temperature, achieving a state called thermal equilibrium.
It’s important to remember that during this exchange, the total energy in the system remains constant. Only its form and distribution change among the water masses. By setting up an equation that reflects this constant energy, we can figure out unknown variables like the mass of water needed to achieve a specific temperature.
Specific Heat Capacity
The specific heat capacity ( \( c \) ) is a property of a material that defines how much heat energy it requires to change its temperature. For water, this value is approximately 4.18 Joules per gram per degree Celsius, but in this exercise, it cancels out due to both masses being in water.
This metric becomes especially useful in problems involving temperature changes because it allows us to calculate how much energy is needed to change the temperature of a given mass.
  • It is constant for a given material, influencing how quickly it heats up or cools down.
  • It tells us that more energy is needed to raise the temperature of a substance with a high specific heat capacity than one with a lower value.
In our exercise, even though we don't directly use the value of water's specific heat capacity, understanding it helps us realize why and how the energy exchange takes place during heating.
Thermal Equilibrium
Thermal equilibrium is reached when two substances in thermal contact no longer transfer heat between each other, meaning they have achieved the same temperature. In our exercise, after adding 1950 grams of boiling water to the colder water, both eventually reach the desired temperature of 75°C.
This concept is vital in determining the end condition of any heat transfer scenario. At equilibrium:
  • The temperature of all involved bodies is the same.
  • No more heat flows between them.
In practical applications, understanding thermal equilibrium is essential for solving problems involving mixed temperatures, such as in cooking or climate control systems. Identifying this stable point allows for calculating how much energy or mass is needed to achieve a desired temperature.

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Most popular questions from this chapter

BIO Conduction Through the Skin. The blood plays an important role in removing heat from the body by bringing this energy directly to the surface where it can radiate away. Nevertheless, this heat must still travel through the skin before it can radiate away. Assume that the blood is brought to the bottom layer of skin at 37.0\(^\circ\)C and that the outer surface of the skin is at 30.0\(^\circ\)C. Skin varies in thickness from 0.50 mm to a few millimeters on the palms and soles, so assume an average thickness of 0.75 mm. A 165-lb, 6-ft-tall person has a surface area of about 2.0 m\(^2\) and loses heat at a net rate of 75 W while resting. On the basis of our assumptions, what is the thermal conductivity of this person’s skin?

CP While painting the top of an antenna 225 m in height, a worker accidentally lets a 1.00-L water bottle fall from his lunchbox. The bottle lands in some bushes at ground level and does not break. If a quantity of heat equal to the magnitude of the change in mechanical energy of the water goes into the water, what is its increase in temperature?

BIO While running, a 70-kg student generates thermal energy at a rate of 1200 W. For the runner to maintain a constant body temperature of 37\(^\circ\)C, this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the energy could not flow out of the student’s body, for what amount of time could a student run before irreversible body damage occurred? (Note: Protein structures in the body are irreversibly damaged if body temperature rises to 44\(^\circ\)C or higher. The specific heat of a typical human body is 3480 J / kg \(\cdot\) K, slightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heats.)

You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped 10.0 K. What is its temperature change in (a) F\(^\circ\) and (b) C\(^\circ\)?

A vessel whose walls are thermally insulated contains 2.40 kg of water and 0.450 kg of ice, all at 0.0\(^\circ\)C. The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to 28.0\(^\circ\)C? You can ignore the heat transferred to the container.

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