Question

When \(155 \mathrm{~mL}\) of water at \(26^{\circ} \mathrm{C}\) is mixed with \(75 \mathrm{~mL}\) of water at \(85^{\circ} \mathrm{C},\) what is the final temperature? (Assume that no heat is released to the surroundings; \(d\) of water \(=1.00 \mathrm{~g} / \mathrm{mL}\).)

Short answer

The final temperature of the water is approximately 45.24°C.

Step-by-step solution

- Identify Given Data

Determine the given quantities: Volume of water at 26°C (V_1)= 155 mL, Volume of water at 85°C (V_2) = 75 mL, Density of water (d) = 1 g/mL.

- Convert Volume to Mass

Since the density of water is 1 g/mL, the masses will be equal to the volumes. Thus, the mass of water at 26°C (M_1) = 155 g and the mass of water at 85°C (M_2) = 75 g.

- Write Heat Transfer Equation

Use the principle of conservation of energy. The heat lost by hotter water equals the heat gained by cooler water.\[ M_1 c (T_f - T_1) = M_2 c (T_2 - T_f) \]where T_f = final temperature, T_1 = initial temperature of cooler water, T_2 = initial temperature of hotter water, and c = specific heat capacity of water (cancels out as it's the same for both).

- Substitute Values

Equation: \[ 155 (T_f - 26) = 75 (85 - T_f) \]

- Solve for Final Temperature

Expand and solve the equation:\[ 155T_f - 4030 = 6375 - 75T_f \]Add 75T_f to both sides:\[ 230T_f - 4030 = 6375 \]Add 4030 to both sides:\[ 230T_f = 10405 \]Divide by 230:\[ T_f \approx 45.24^{\text{o}}C \]

Key concepts

heat transfer

Understanding how heat transfer works is essential in solving problems involving mixing liquids of different temperatures. Heat transfer is the movement of thermal energy from one object to another. When two bodies at different temperatures come into contact, heat flows from the hotter body to the cooler body until thermal equilibrium is reached. This results in a final uniform temperature of both bodies.
In our exercise, two volumes of water at different temperatures are mixed. The heat energy from the hotter water is transferred to the cooler water. This process continues until both volumes of water reach the same temperature. In simpler terms, the cooler water gains heat, and the hotter water loses heat, equalizing their temperatures over time.
It's also important to note that during this process, no heat is lost to the surroundings, meaning the total thermal energy in the system remains constant.

conservation of energy

The principle of conservation of energy is pivotal in understanding the final temperature calculation. This principle states that energy cannot be created or destroyed, only transferred or transformed from one form to another. In terms of heat transfer, the energy lost by the hotter substance is exactly equal to the energy gained by the cooler substance.
To apply this to our water mixing problem, we set up an equation based on the conservation of energy. The heat lost by the hot water (Equation: \[M_2 c (T_2 - T_f)\]) equals the heat gained by the cold water (Equation: \[M_1 c (T_f - T_1)\]). The specific heat capacity of water (c) is the same for both quantities of water and cancels out, simplifying our calculations. By substituting the known values into the equation, we find the final temperature where the energy of the system is balanced.

specific heat capacity

Specific heat capacity is another critical concept in this exercise. It is defined as the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. For water, this value is approximately 4.18 Joules per gram per degree Celsius.
In our problem, since both quantities of water are the same substance, their specific heat capacities are the same. This allows us to cancel out the specific heat capacity when we write our equation for heat transfer, making our calculations simpler.
Essentially, knowing the specific heat capacity helps us understand how much energy is required to change the temperature of the water and supports the calculations we perform when solving for the final temperature.