Question

How many \(\mathrm{mL}\) of water at \(10^{\circ} \mathrm{C}\) ( 2 significant figures) must be added to \(75 \mathrm{~mL}\) of water at \(35^{\circ} \mathrm{C}\) to obtain a final temperature of \(19^{\circ} \mathrm{C} ?\) (Make the same assumptions as in Question 9.)

Short answer

Answer: 130 mL

Step-by-step solution

Write the heat transfer equation

Since we know that the heat lost by the hot water is equal to the heat gained by the cold water, we can express this relationship as: \(Q_{hot} = Q_{cold}\) where - \(Q_{hot}\) is the heat lost by the hot water, - \(Q_{cold}\) is the heat gained by the cold water.

Substitute the heat equation in terms of mass, specific heat, and temperature change

We know that the heat equation is given by \(Q = mc\Delta T\), where \(m\) is the mass of the substance, \(c\) is its specific heat, and \(\Delta T\) is the change in temperature. Since the mass of a substance is equal to its volume multiplied by its density, we can rewrite the equation in terms of volume as: \(V_{hot} \rho c (\Delta T)_{hot} = V_{cold} \rho c (\Delta T)_{cold}\) Here, \(V_{hot}\) and \(V_{cold}\) are the volumes of the hot and cold water, \(\rho\) is the density of water, and \((\Delta T)_{hot}\) and \((\Delta T)_{cold}\) are the temperature changes for the hot and cold water. Since the density and specific heat of water are constant, they can be eliminated from the equation: \(V_{hot} (\Delta T)_{hot} = V_{cold} (\Delta T)_{cold}\)

Calculate the temperature changes for hot and cold water

The change in temperature for the hot water \((\Delta T)_{hot}\) is given by: \((\Delta T)_{hot} = T_{initial, hot} - T_{final}\) Similarly, the change in temperature for the cold water \((\Delta T)_{cold}\) is given by: \((\Delta T)_{cold} = T_{final} - T_{initial, cold}\) Substitute the given temperatures into the equations: \((\Delta T)_{hot} = 35^{\circ} \mathrm{C} - 19^{\circ} \mathrm{C} = 16^{\circ} \mathrm{C} \) \((\Delta T)_{cold} = 19^{\circ} \mathrm{C} - 10^{\circ} \mathrm{C} = 9^{\circ} \mathrm{C} \)

Substitute the known values into the equation and solve for the unknown volume

We have: \(V_{hot} (\Delta T)_{hot} = V_{cold} (\Delta T)_{cold}\) Substitute the known values: \((75 \mathrm{~mL})(16^{\circ} \mathrm{C}) = V_{cold} (9^{\circ} \mathrm{C})\) Now, solve for \(V_{cold}\): \(V_{cold} = \frac{(75 \mathrm{~mL})(16^{\circ} \mathrm{C})}{(9^{\circ} \mathrm{C})} = 133.3 \mathrm{~mL}\) (rounded to one decimal place) Since we want the answer to 2 significant figures, the volume of water at \(10^{\circ} \mathrm{C}\) to be added is: \(V_{cold} = 130 \mathrm{~mL}\)

Key concepts

Thermodynamics and Heat Transfer

When studying thermodynamics in the context of chemistry, we are often concerned with how heat energy is transferred between substances. Heat transfer is a fundamental concept in thermodynamics that explains the exchange of heat energy during physical and chemical processes. The principle that guides this exchange is known as the First Law of Thermodynamics, which states that energy cannot be created or destroyed, only transformed from one form to another or transferred.

In chemistry problems, we commonly encounter scenarios where heat is transferred from a warmer substance to a cooler one, until thermal equilibrium is reached. This transfer continues until both substances have the same temperature, with the warmer substance cooling down and the cooler substance warming up. The task often requires us to calculate the amount of heat gained or lost by the substances involved, which is a direct application of understanding thermodynamics and the conservation of energy principle.

Specific Heat Capacity

Understanding specific heat capacity is crucial in solving many heat transfer problems. It is defined as the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin). Specific heat capacity, often symbolized as c, is a unique value for each substance, indicating its thermal inertia, or how resistant it is to changes in temperature.

The knowledge of a substance's specific heat capacity allows us to calculate the heat exchange when a substance undergoes a temperature change. In the context of the exercise provided, water has a specific heat capacity of approximately 4.18 joules per gram per degree Celsius (\(4.18 \frac{J}{g \times ^{\text{o}}C}\)). This knowledge is essential when we set up our heat transfer equations, as it ensures the accuracy of our temperature change calculations.

Temperature Change Calculation

The temperature change calculation is an application of the concept of heat transfer. We use the formula Q = mcΔT, where Q is the heat transfer in joules, m is the mass of the substance in grams, c is the specific heat capacity, and ΔT (Delta T) is the temperature change in degrees Celsius.

To improve on the student’s understanding, let's break down the process of solving the exercise with an emphasis on the practical application of the formula. In our exercise, hot and cold water are mixed, and we aim to find out how much cold water is needed to reach a desired final temperature. The key is to realize that the heat lost by the hot water is gained by the cold water. Therefore, we equate the heat lost and gained and rearrange the formula to solve for the unknown volume of cold water. This process shows the conservation of energy in practice and enforces the importance of precise calculation in temperature changes for various thermodynamics applications.