Question

When $225\mathrm{mL}$ of ${\mathrm{H}}_2\mathrm{O}$ at ${25}^{\circ }\mathrm{C}$ are mixed with $85\mathrm{mL}$ of water at ${89}^{\circ }\mathrm{C}$, what is the final temperature? (Assume that no heat is lost to the surroundings; $d_{{\mathrm{H}}_2\mathrm{O}}=1.00\mathrm{g}/\mathrm{mL}$.)

Short answer

Answer: The final temperature of the mixture is approximately 46.29°C.

Step-by-step solution

Calculate the mass of each water sample

Calculate the mass of each water sample by multiplying the density of water with the volume of each sample. Mass of first water sample: $m_1=d_{H_{2}O}\times V_1=1.00\frac{g}{mL}\times 225mL=225g$ Mass of second water sample: $m_2=d_{H_{2}O}\times V_2=1.00\frac{g}{mL}\times 85mL=85g$

Write the heat transfer equation for both samples

Since the heat gained by the colder water is equal to the heat lost by the hotter water, we can write the equation: $q_1=q_2$ Where $q_1$ is the heat gained by the first sample and $q_2$ is the heat lost by the second sample.

Use the heat transfer formula and substitute values

We have $q=mc\Delta T$, where $c$ is the specific heat capacity of water, which is $4.18J/(g{\cdot }^{\circ }C)$. Let $T_f$ be the final temperature of the mixture; then, substitute the values in the equation we derived in Step 2: $m_{1}c(T_f-T_1)=m_{2}c(T_2-T_f)$

Solve for the final temperature $T_f$

Substitute the values of the mass, specific heat capacities, and initial temperatures in the equation, and solve for $T_f$: $225g\cdot 4.18\frac{J}{g{\cdot }^{\circ }C}(T_f-{25}^{\circ }C)=85g\cdot 4.18\frac{J}{g{\cdot }^{\circ }C}({89}^{\circ }C-T_f)$ The specific heat capacity cancels out, so we get: $225(T_f-25)=85(89-T_f)$ Now, solve for $T_f$: $225T_f-5625=85\cdot 89-85T_f$ $225T_f+85T_f=5625+85\cdot 89$ $310T_f=14350$ $T_f=\frac{14350}{310}={46.29}^{\circ }C$ The final temperature of the mixture is approximately ${46.29}^{\circ }C$.

Key concepts

Specific Heat Capacity

Imagine you're holding a hot cup of coffee. The warmth you feel is a result of energy, in our context, heat energy, that's being transferred from the coffee to your hands. This process is central to understanding how heat moves from one object to another and is described using a term called 'specific heat capacity'.

Specific heat capacity, denoted by the symbol 'c', refers to the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 Kelvin). It is crucial for calculating how substances will respond when they exchange heat. This property varies from substance to substance; for instance, water has a high specific heat capacity of 4.18 Joules per gram per degree Celsius ($4.18\frac{J}{g\text{textdegree}C}$). This means that water needs a lot of heat energy to increase in temperature, which is why it's so effective at absorbing and transferring heat in our homes and the environment.

Understanding the specific heat capacity is important in many applications, from engineering to environmental science to culinary arts. It helps us predict how long it will take for an object to heat up or cool down. This concept is of particular importance in thermochemistry calculations, where it’s used to determine the heat transfer during chemical reactions, phase changes, or even simple processes like mixing two quantities of water at different temperatures, as seen in our exercise.

Thermochemistry Calculations

Thermochemistry calculations go hand-in-hand with specific heat capacity when we're working out how to calculate the exchange of energy as heat. These calculations typically involve the use of formulas and the conservation of energy principle, which states that energy cannot be created or destroyed, only transformed.

One fundamental formula is the heat transfer equation: $q=mc\mathrm{\Delta }T$ where 'q' represents the heat energy transferred, 'm' is the mass of the substance, 'c' is its specific heat capacity, and $\mathrm{\Delta }T$ is the change in temperature. In our exercise, we're applying this formula to find the equilibrium temperature when two water samples at different temperatures come in contact.

To perform thermochemistry calculations, you need to set the heat gained by one body equal to the heat lost by another, as they will reach thermal equilibrium. This relationship allows you to solve for unknown quantities, such as the final temperature of a mixture. The power of thermochemistry calculations lies in their ability to paint a quantitative picture of thermal processes, translating qualitative observations into meaningful data.

Temperature Change

We encounter temperature change in everyday life, from cooking food on the stove to feeling the breezes of changing seasons. In science, and particularly chemistry, understanding temperature change is essential to exploring how energy is exchanged during reactions or between substances.

Temperature change, denoted by $\mathrm{\Delta }T$, is the difference between the final and initial temperatures of a system and can be positive or negative depending on whether the system is gaining or losing heat. In the context of the exercise, mixing two water samples with different initial temperatures results in a temperature change leading to a new equilibrium temperature.

Exploring temperature change provides insight into endothermic and exothermic processes, and is closely related to heat capacity, as it helps determine how much a substance's temperature will change when a certain amount of heat is added or removed. The final temperature in our exercise – approximately ${46.29}^{\circ }C$ – is a direct result of the heat transfer and the capacities of the individual water samples to absorb or release heat.