Question

A \(50.0-\mathrm{g}\) sample of water at \(100 .\) ' \(\mathrm{C}\) is poured into a \(50.0-\mathrm{g}\) 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 mixture will be \(62.5^{\circ}\,\mathrm{C}\).

Step-by-step solution

Identify given values and formula

We are given: - Mass of water sample 1: \(m_1 = 50.0 \,\mathrm{g}\) - Temperature of water sample 1: \(T_{1i} = 100^{\circ}\,\mathrm{C}\) - Mass of water sample 2: \(m_2 = 50.0\, \mathrm{g}\) - Temperature of water sample 2: \(T_{2i} = 25^{\circ}\,\mathrm{C}\) Our goal is to find the final temperature (\(T_f\)) of the water mixture. The heat exchange formula is: \(q = mc \Delta T\) Where: - q is the heat exchanged, - m is the mass of the substance, - c is the specific heat capacity (for water: \(c = 4.18 \, \mathrm{J / (g \cdot K)}\)), - \(\Delta T\) is the change in temperature.

Set up the heat exchange equation

The heat gained by water sample 1 (the colder water) is equal to the heat lost by water sample 2 (the hotter water). So, we can set up the following equation: \(m_1c(T_f - T_{1i}) = m_2c(T_{2i} - T_f)\) We can plug in the known values: \(50.0 \cdot 4.18 (T_f - 25) = 50.0 \cdot 4.18 (100 - T_f)\) Notice that the mass and specific heat capacity of the water are the same for both samples, so we can cancel out these terms: \(T_f - 25 = 100 - T_f\)

Solve for the final temperature

Now we can solve for \(T_f\): \(2T_f = 100 + 25\) \(T_f = 62.5^{\circ}\,\mathrm{C}\)

State the final answer

The final temperature of the water mixture will be \(62.5^{\circ}\,\mathrm{C}\).

Key concepts

Specific Heat Capacity

Understanding the concept of specific heat capacity is fundamental in explaining how different substances respond to energy changes in the form of heat. 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). In mathematical terms, it's represented as the 'c' in the heat exchange formula:
\( q = mc\Delta T \).

This property is unique to each substance and reflects its capacity to store thermal energy. Water, for instance, has a relatively high specific heat capacity (about 4.18 J/g•K), which means it can hold more heat energy per gram for each degree rise in temperature compared to many other substances. This is why water is often used as a coolant; it has the ability to absorb significant amounts of heat before it shows substantial changes in temperature.

Temperature Change Calculation

Calculating temperature change is critical in various fields, particularly chemistry and physics. When dealing with heat exchange, it's essential to understand how to compute the change in temperature as a result of heat transfer. The change in temperature, denoted as \(\Delta T\), is found by using the formula:
\( \Delta T = T_f - T_i \),

where \(T_i\) and \(T_f\) are the initial and final temperatures, respectively. In our exercise, this formula allows us to determine how much the temperature of water needs to change to reach thermal equilibrium. When two bodies at different temperatures come into contact, heat will flow from the warmer body to the cooler one until the temperatures equilibrate – this is the crux of temperature change calculation.

Thermal Equilibrium

The ultimate goal in our exercise is to achieve thermal equilibrium, a state where no net heat flows between the substances in contact. This concept is established on the Zeroth Law of Thermodynamics, which states that if two systems are in thermal equilibrium with a third system, they must be in thermal equilibrium with each other. It's a guiding principle for many thermodynamics problems.

In practical terms, when two samples of water at different temperatures are mixed, as in the exercise, they will exchange heat until both reach the same temperature – the final temperature. Heat exchange continues until the energy lost by the hotter substance equals the energy gained by the cooler one. The equation reflecting equal heat exchange,
\( m_1c(T_f - T_{1i}) = m_2c(T_{2i} - T_f) \),
set up in Step 2 of the solution, is an expression of this principle. Once we solve for the final temperature using this relationship, we have identified the point at thermal equilibrium – no more net heat transfer occurs after this point.