Question

What is the final temperature when \(70.0 \mathrm{~g}\) of water at \(80.0^{\circ} \mathrm{C}\) is mixed with \(30.0 \mathrm{~g}\) of water at \(25.0^{\circ} \mathrm{C}\) ?

Short answer

The final temperature when 70.0 g of water at 80.0°C is mixed with 30.0 g of water at 25.0°C will be approximately 62.14 °C.

Step-by-step solution

Determine Initial Values

Our two quantities of water are 70g at 80 degrees Celsius and 30g at 25 degrees Celsius. The specific heat capacity of water is roughly \(4.18 \, \mathrm{J/g\cdot °C}\). The water that's starting at 80 degrees is going to cool, while the water that's starting at 25 degrees is going to heat up.

Set Up Equations

Set up the heat transfer equation: the heat lost by warmer water equals heat gained by cooler water. It can be written as: \((mass_1) \cdot (specific \, heat) \cdot (T_{final} - T_{initial1}) = (mass_2) \cdot (specific \, heat) \cdot (T_{initial2} - T_{final})\). Here, \(T_{initial1} = 80 °C, T_{initial2} = 25 °C\) are initial temperatures and \(T_{final}\) is the final temperature we want to find.

Simplify and Solve

Simplify the equation by cancelling out the common term, which is 'Specific Heat'. The result is \(70 \cdot (T_{final} - 80) = 30 \cdot (25 - T_{final})\). Solve this equation to find \(T_{final}\). This is a simple linear equation and can be solved by first distributing the numbers on both sides, then bringing all terms involving \(T_{final}\) to one side and constants on the other, and finally simplifying to determine \(T_{final}\).

Calculate Final Temperature

Solving the equation from Step 3 gives \(T_{final}\) as approximately 62.14 °C.

Key concepts

Specific Heat Capacity

Specific heat capacity is a crucial concept in thermodynamics. It defines how much heat energy is required to change the temperature of a substance. For water, its specific heat capacity is approximately \(4.18 \, \mathrm{J/g \cdot °C}\). This means that to raise the temperature of 1 gram of water by 1 degree Celsius, you need 4.18 joules of energy.
Specific heat capacity helps us understand why different substances heat up or cool down at different rates. Substances with a high specific heat capacity, like water, can absorb a lot of heat without a significant change in temperature. This property makes water excellent for thermal management and energy storage.
When solving problems involving specific heat, always remember to consider the mass of the substance and the temperature change involved. The formula to calculate heat transfer using specific heat is:
\[ q = m \cdot c \cdot \Delta T \]
where:

• \(q\) is the heat energy
• \(m\) is the mass
• \(c\) is the specific heat capacity
• \(\Delta T\) is the change in temperature

Heat Transfer

Heat transfer is the movement of thermal energy from one object or substance to another. When two bodies at different temperatures come into contact, heat transfer occurs. As a rule, heat flows from the warmer body to the cooler one until thermal equilibrium is reached.
In our exercise, we have two masses of water at different initial temperatures. The water at \(80^{\circ} \mathrm{C}\) will transfer some of its thermal energy to the water at \(25^{\circ} \mathrm{C}\), causing the hot water to cool down and the cold water to warm up.
The principle of heat exchange states that the heat lost by the hot object (heat going out) equals the heat gained by the cold object (heat coming in). Thus, we can set up a heat transfer equation:
\[ m_1 \cdot c \cdot (T_{final} - T_{initial1}) = m_2 \cdot c \cdot (T_{initial2} - T_{final}) \]
where:

• \(m_1\) and \(m_2\) are the masses of the two water samples
• \(T_{initial1}\) and \(T_{initial2}\) are their respective initial temperatures
• \(T_{final}\) represents the shared final temperature

Using this understanding of heat transfer can solve many practical problems regarding temperature changes in different systems.

Temperature Calculation

Temperature calculation involves finding the final temperature (\(T_{final}\)) when substances exchange heat. In heat transfer problems, we solve the heat balance equation to find \(T_{final}\). This equation is derived from the principle that the heat lost by the warmer substance equals the heat gained by the cooler substance.
To solve such problems, we follow a few key steps:

• Identify the initial temperatures and masses of both substances involved.
• Set up the heat transfer equation: \(m_1 \cdot (T_{final} - T_{initial1}) = m_2 \cdot (T_{initial2} - T_{final})\).
• Simplify the equation by eliminating common terms, such as specific heat if it's the same for both substances.
• Solve the equation for \(T_{final}\).

For example, in our problem with 70g of water at \(80^{\circ} \mathrm{C}\) and 30g of water at \(25^{\circ} \mathrm{C}\):
- The equation becomes \(70 \cdot (T_{final} - 80) = 30 \cdot (25 - T_{final})\).
- Distribute and rearrange terms to isolate \(T_{final}\).
- Solve to find \(T_{final} \approx 62.14^{\circ} \mathrm{C}\).
This calculation method ensures accurate results in problems involving heat exchange and temperature changes.