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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

A 500.0-g chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing 1.00 kg of water at room temperature (20.0\(^\circ\)C). After waiting and gently stirring for 5.00 minutes, you observe that the water’s temperature has reached a constant value of 22.0\(^\circ\)C. (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? (b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain. (c) If the heat absorbed by the Styrofoam actually is not negligible, how would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.

A copper calorimeter can with mass 0.100 kg contains 0.160 kg of water and 0.0180 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead at 255\(^\circ\)C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

In very cold weather a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is -20\(^\circ\)C, what amount of heat is needed to warm to body temperature (37\(^\circ\)C) the 0.50 L of air exchanged with each breath? Assume that the specific heat of air is 1020 J / kg \(\cdot\) K and that 1.0 L of air has mass \(1.3 \times 10{^-}{^3} kg\). (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

To measure the specific heat in the liquid phase of a newly developed cryoprotectant, you place a sample of the new cryoprotectant in contact with a cold plate until the solution's temperature drops from room temperature to its freezing point. Then you measure the heat transferred to the cold plate. If the system isn't sufficiently isolated from its room-temperature surroundings, what will be the effect on the measurement of the specific heat? (a) The measured specific heat will be greater than the actual specific heat; (b) the measured specific heat will be less than the actual specific heat; (c) there will be no effect because the thermal conductivity of the cryoprotectant is so low; (d) there will be no effect on the specific heat, but the temperature of the freezing point will change.

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

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