# An unknown volume of water at 18.2°C is added to 33.5 mL of water at 35.0°C. If the final temperature is 23.5°C, what was the unknown volume?

Try to predict what you would expect the answer to be before performing any calculations.

It is noticeable that the final temperature is more in line with the colder water sample's temperature than the warmer sample's temperature.

As a result, you should anticipate that the colder sample will have a larger volume than the warmer sample.

The mass of the warmer water sample must now be calculated using the density of water, which can be found here for a range of temperatures.

Water Density (https://tutor.hix.ai)

The equation that shows a connection between heat gained or lost and temperature change will now be your preferred tool.

The fact that the heat gained by the colder sample will equal the heat lost by the warmer sample holds the key to solving this puzzle.

Heat loss also has a negative sign, so the minus sign is required.

The response, rounded to three sig figs, is

The volume of the colder sample was larger than the volume of the warmer sample, proving that the first prediction was accurate.

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To find the unknown volume of water added, you can use the principle of conservation of energy, specifically the equation:

[ q_{\text{lost}} + q_{\text{gain}} = 0 ]

Where ( q_{\text{lost}} ) represents the heat lost by the hot water and ( q_{\text{gain}} ) represents the heat gained by the cold water. This equation can be written as:

[ m_1 \cdot c \cdot (T_f - T_1) + m_2 \cdot c \cdot (T_f - T_2) = 0 ]

Where:

- ( m_1 ) is the mass of the hot water (given),
- ( m_2 ) is the mass of the cold water (unknown),
- ( c ) is the specific heat capacity of water,
- ( T_f ) is the final temperature (given),
- ( T_1 ) is the initial temperature of the hot water (given), and
- ( T_2 ) is the initial temperature of the cold water (given).

Rearranging the equation to solve for ( m_2 ), you get:

[ m_2 = - \frac{m_1 \cdot c \cdot (T_f - T_1)}{c \cdot (T_f - T_2)} ]

Now, plug in the given values and solve for ( m_2 ), which represents the mass of the cold water. Finally, convert the mass of the cold water to volume using the density of water.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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