Question

A 30.0 -g sample of water at \(280 . \mathrm{K}\) is mixed with 50.0 g water at \(330 . \mathrm{K}\) . Calculate the final temperature of the mixture assuming no heat loss to the surroundings.

Short answer

The final temperature of the water mixture is \(311.25\:K\).

Step-by-step solution

Write the given information

Mass of water sample 1, \(m_1 = 30.0\:g\) Initial temperature of water sample 1, \(T_1 = 280\:K\) Mass of water sample 2, \(m_2 = 50.0\:g\) Initial temperature of water sample 2, \(T_2 = 330\:K\) The specific heat capacity of water, \(c = 4.184\:J/(g\:K)\)

Calculate the heat transfer for both water samples

Let's assume the final temperature of the mixture is \(T_f\). For water sample 1, the heat transfer will be: \(Q_1 = m_1c\Delta T_1 = m_1c(T_f - T_1)\) For water sample 2, the heat transfer will be: \(Q_2 = m_2c\Delta T_2 = m_2c(T_2 - T_f)\) Since heat lost by water sample 2 is equal to the heat gained by water sample 1 and there's no heat loss to surroundings, we can write the equation as: \(Q_1 = Q_2\)

Solve for the final temperature

Plug in the values of \(Q_1\) and \(Q_2\) in the equation \(Q_1 = Q_2\): \(m_1c(T_f - T_1) = m_2c(T_2 - T_f)\) Divide both sides by \(c\): \(m_1(T_f - T_1) = m_2(T_2 - T_f)\) Now, solve for \(T_f\): \(T_f(m_1 + m_2) = m_1T_1 + m_2T_2\) \(T_f = \frac{m_1T_1 + m_2T_2}{m_1 + m_2}\) Plug in the values: \(T_f = \frac{(30.0\:g)(280\:K) + (50.0\:g)(330\:K)}{30.0\:g + 50.0\:g}\) \(T_f = \frac{8400 + 16500}{80}\) \(T_f = \frac{24900}{80}\) \(T_f = 311.25\:K\)

State the final temperature

The final temperature of the water mixture is \(311.25\:K\).

Key concepts

Specific Heat Capacity

Specific heat capacity is an important concept when discussing heat transfer between substances. It refers to the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius or Kelvin. For water, this value is known to be relatively high at 4.184 J/(g K).
This implies that water can absorb a significant amount of heat before its temperature rises considerably. In practical terms, understanding specific heat capacity helps explain why water is effective at storing heat, making it useful in various applications like heating systems.

• It's a measure of heat storage capacity.
• Varies between different substances, affecting how they react to temperature changes.

By knowing the specific heat capacity, we can calculate the heat required to change a substance's temperature, which is crucial for solving calorimetry problems.

Thermal Equilibrium

Thermal equilibrium occurs when two or more substances in contact no longer exchange heat. At this point, these substances reach the same temperature and stop undergoing changes in energy.
In the context of our exercise, when you mix two samples of water at different temperatures, they transfer heat until they reach thermal equilibrium which is their final common temperature. This concept is vital in understanding how heat is conserved and distributed between objects.

• Heat moves from a hotter object to a cooler one until balanced.
• No further heat exchange once equilibrium is reached.

This process helps in the calculation of the final temperature when mixing substances of different initial temperatures, assuming no heat is lost to the surroundings.

Calorimetry

Calorimetry is the science of measuring the transfer of heat during chemical reactions or physical changes. It is primarily used to determine the heat capacity of substances.
This method involves using equations and known properties, such as specific heat capacity, to calculate the heat involved in a process. In our exercise, calorimetry helps us determine the final temperature when two water samples at different temperatures are mixed:

• It uses the equation: \(Q = mc\Delta T\)
• Q is the heat exchanged, m is the mass, c is specific heat capacity, and \(\Delta T\) is the temperature change.

By setting the heat lost by the warmer sample equal to the heat gained by the cooler one, calorimetry allows us to solve for unknown variables like the final temperature.