Chapter 17: Problem 34
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.
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.
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.
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:
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.