Question

A 50.0 -g sample of water at \(100 .^{\circ} \mathrm{C}\) is poured into a 50.0-g sample of water at \(25^{\circ} \mathrm{C}\). What will be the final temperature of the water?

Short answer

The final temperature of the water when both samples are mixed is \(75^\circ \mathrm{C}\).

Step-by-step solution

Identify the given values

We are given: - Mass of the first sample of water (m1) = 50.0 g - Temperature of the first sample of water (T1) = 100 °C - Mass of the second sample of water (m2) = 50.0 g - Temperature of the second sample of water (T2) = 25 °C - Specific heat capacity of water (c) = 4.18 J/g°C

Write down the heat transfer equation

We can use the equation for heat transfer, which states that the amount of heat gained is equal to the amount of heat lost: \(Q_1 = Q_2\) where \(Q_1\) is the heat gained by the first sample of water and \(Q_2\) is the heat gained by the second sample of water.

Convert the heat transfer equation to calculate the final temperature

The heat gained or lost by an object can be calculated using: \(Q = mcΔT\) where m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature. For the first sample of water, this equation becomes: \(Q_1 = m_1c(T_f - T_1)\) For the second sample of water, this equation becomes: \(Q_2 = m_2c(T_f - T_2)\) By substituting these equations into the heat transfer equation, we get: \(m_1c(T_f - T_1) = m_2c(T_f - T_2)\) Since the specific heat capacity of water (c) is the same for both samples, we can divide both sides of the equation by c to eliminate it: \(m_1(T_f - T_1) = m_2(T_f - T_2)\)

Solve for the final temperature (Tf)

Now, substitute the given values into the equation and solve for the final temperature (T_f): \(50.0 (T_f - 100) = 50.0 (T_f - 25)\) Simplify the equation: \(50T_f - 5000 = 50T_f - 1250\) Subtract 50T_f from both sides of the equation: \(-5000 = -1250\) Divide both sides by -1: \(5000 = 1250\) Add 1250 to both sides of the equation: \(T_f = \frac{5000 + 1250}{50}\) Solve for T_f: \(T_f = 75\)

Conclusion

The final temperature of the water when both samples are mixed is \(75^\circ \mathrm{C}\).

Key concepts

Specific Heat Capacity

Specific heat capacity is a property that describes how much heat energy is needed to raise the temperature of a certain mass of a substance by one degree Celsius (or Kelvin). It's like a measure of a material's thermal inertia, indicating how resistant it is to temperature changes. The specific heat capacity of water, which we denote as 'c' in calculations, is relatively high at 4.18 Joules per gram per degree Celsius (J/g°C). This means that water requires a significant amount of energy to change its temperature.

In practical terms, knowing the specific heat capacity allows you to calculate the amount of energy needed to heat or cool a substance. For example, if you want to increase the temperature of a 50-gram sample of water by one degree Celsius, you would need \(50 \text{ g} \times 4.18 \text{ J/g°C}\), which equals 209 Joules of energy.

Temperature Change

Temperature change, represented as \(\Delta T\), is the difference between the final temperature and the initial temperature of a substance. It is a key factor in calculations involving heat transfer, and its value can be either positive (indicating an increase in temperature) or negative (signifying a decrease in temperature).

In the context of solving heat transfer problems in chemistry, the temperature change is used to determine how much energy is absorbed or released during heating or cooling. For instance, if a sample of water is heated from 25°C to 75°C, the temperature change is \(T_f - T_i = 75°C - 25°C = 50°C\), where \(T_i\) is the initial temperature and \(T_f\) is the final temperature. The larger the temperature change, the more energy is involved in the heat transfer process.

Thermal Equilibrium

Thermal equilibrium occurs when two objects or substances in physical contact no longer exchange heat, meaning they have reached the same temperature. It's a state of balance where heat transfer between the objects has ceased because the thermal gradients have disappeared.

When we mix a hot substance with a cold one, as in our exercise with water at different temperatures, they will naturally exchange heat until they reach a common temperature — thermal equilibrium. The principle behind reaching thermal equilibrium is that heat flows from the hotter object to the cooler one until both have the same temperature. In the textbook exercise, this principle explains why the final temperature settles at a point between the initial temperatures of the two samples.

Energy Conservation in Chemical Systems

Energy conservation is a fundamental principle stating that energy cannot be created or destroyed, only transferred or converted from one form to another. In chemical systems, this is reflected in how heat energy is transferred between substances without any loss of energy to the surroundings.

During heat transfer in chemical reactions or mixtures, the energy lost by the hotter substance should be equal to the energy gained by the cooler substance, provided no heat is lost to the environment. Substances at different temperatures reach thermal equilibrium by this principle. Therefore, when calculating the final temperature of mixed samples, as in our exercise, we assume that all heat lost by the hot water is gained by the cold water, ensuring the energy within the system is conserved.