Question

An unknown volume of water at \(18.2^{\circ} \mathrm{C}\) is added to \(24.4 \mathrm{~mL}\) of water at \(35.0^{\circ} \mathrm{C}\). If the final temperature is \(23.5^{\circ} \mathrm{C},\) what was the unknown volume? (Assume that no heat is released to the surroundings; \(d\) of water \(=1.00 \mathrm{~g} / \mathrm{mL} .\)

Short answer

Approximately 53 mL.

Step-by-step solution

Understand the Heat Transfer

Since no heat is released to the surroundings, the heat lost by the warmer water will be equal to the heat gained by the cooler water. This can be expressed as: \( q_{\text{lost}} = q_{\text{gained}} \)

Use the Heat Transfer Formula

The formula for heat transfer is: \( q = m \times c \times \triangle T \) where \( m \) is the mass, \( c \) is the specific heat capacity (\( 4.18 \frac{J}{g \times ^\text{circ} C} \) for water), and \( \triangle T \) is the change in temperature. Since density \( d = 1.00 \frac{g}{mL} \), the mass \( m \) in grams is equal to the volume in mL.

Set up the Equations

For the warmer water: \[ q_{\text{lost}} = (24.4 \text{ mL} \times 1.00 \frac{g}{mL}) \times 4.18 \frac{J}{g \times ^\text{circ} C} \times (35.0 - 23.5)^{\text{circ} C} \]For the cooler water: \[ q_{\text{gained}} = (V_{\text{unknown}} \times 1.00 \frac{g}{mL}) \times 4.18 \frac{J}{g \times ^\text{circ} C} \times (23.5 - 18.2)^{\text{circ} C} \]

Simplify the Equations

Calculate the heat lost by the warmer water: \[ q_{\text{lost}} = 24.4 \times 4.18 \times 11.5 \]Calculate the heat gained by the cooler water: \[ q_{\text{gained}} = V_{\text{unknown}} \times 4.18 \times 5.3 \]

Solve for the Unknown Volume

Equate the heat lost to the heat gained: \[ 24.4 \times 4.18 \times 11.5 = V_{\text{unknown}} \times 4.18 \times 5.3 \]Cancel out \( 4.18 \) on both sides and solve for \( V_{\text{unknown}} \): \[ 24.4 \times 11.5 = V_{\text{unknown}} \times 5.3 \]\[ V_{\text{unknown}} = \frac{24.4 \times 11.5}{5.3} \]\[ V_{\text{unknown}} \text{ is approximately } 53 \text{ mL} \]

Key concepts

specific heat capacity

Specific heat capacity measures how much heat energy is needed to change the temperature of a given substance. For water, it is defined as 4.18 Joules per gram per degree Celsius (J/g°C). This means that for every gram of water, 4.18 J are needed to raise or lower its temperature by 1°C. Understanding this is essential when calculating heat transfer in water mixtures.

mass and volume relationship

In this exercise, understanding the relationship between mass and volume is key. Since the density of water is 1.00 g/mL, this implies that 1 mL of water has a mass of 1 gram. Therefore, if you have 24.4 mL of water, its mass is also 24.4 grams. This simple relationship helps in converting volume to mass, simplifying calculations related to heat transfer.

temperature change calculation

Knowing how to calculate temperature change is fundamental. The formula for heat transfer, given by \( q = m \times c \times \Delta T \) where:

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

This means the heat gained or lost by a substance can be calculated knowing its mass, specific heat capacity, and change in temperature. In the given problem, you'll need to calculate separate \( q \) values for the warmer and cooler water, and then set these values equal to each other to find the unknown volume.

density explanation

Density is a measure of how much mass is contained in a given volume. For water, the density is 1.00 g/mL. This is crucial because it allows conversion between volume (in mL) and mass (in grams). Hence, if you have, say, 50 mL of water, its mass would be 50 grams. This constant density provides a simplified method for heat transfer calculations.