Question

For each of the following situations a-c, use choices i-iii to complete the statement: "The final temperature of the water should be.." i. between \(50^{\circ} \mathrm{C}\) and \(90^{\circ} \mathrm{C}\) . ii. \(50^{\circ} \mathrm{C}\) . iii. between \(10^{\circ} \mathrm{C}\) and \(50^{\circ} \mathrm{C}\) . a. 100.0 -g sample of water at \(90^{\circ} \mathrm{C}\) is added to a 100.0 -g sample of water at \(10^{\circ} \mathrm{C}\) . b. A 100.0 -g sample of water at \(90^{\circ} \mathrm{C}\) is added to a \(500.0 . \mathrm{g}\) sample of water at \(10^{\circ} \mathrm{C} .\) c. You have a Styrofoam cup with 50.0 \(\mathrm{g}\) of water at \(10^{\circ} \mathrm{C}\) . You add a 50.0 -g iron ball at \(90^{\circ} \mathrm{C}\) to the water.

Short answer

a. ii. $50^{\circ} \mathrm{C}$. b. iii. between $10^{\circ} \mathrm{C}$ and $50^{\circ} \mathrm{C}$. c. iii. between $10^{\circ} \mathrm{C}$ and $50^{\circ} \mathrm{C}$.

Step-by-step solution

We need to find the final temperature of the water in each situation. To do this, we'll use the concept of heat transfer, which states that the heat gained by a colder substance is equal to the heat lost by a warmer substance. Mathematically, we can express this as: \(Q_{gained} = Q_{lost}\) ##Step 2: Applying the concept to situation a##

In situation a, we have 100.0 g of water at \(90^{\circ} \mathrm{C}\) added to another 100.0 g sample of water at \(10^{\circ} \mathrm{C}\). To find the final temperature, we can set up the equation as: \(m_1c(T_f - T_{1}) = m_2c(T_{2} - T_f)\) where \(m_1\) and \(m_2\) are the masses of the water samples, \(c\) is the specific heat of water, \(T_{1}\) and \(T_{2}\) are the initial temperatures, and \(T_f\) is the final temperature. Since the masses and specific heat are equal in this case, we can simplify the equation to: \(T_f - T_{1} = T_{2} - T_f\) Solving for \(T_f\), we get: \(T_f = \frac{T_1+T_2}{2}\) Plugging in the given values, we find the final temperature: \(T_f = \frac{90 + 10}{2} = 50^{\circ} \mathrm{C}\) Thus, for situation a, the final temperature of the water should be \(50^{\circ} \mathrm{C}\) (ii). ##Step 3: Applying the concept to situation b##

In situation b, we have 100.0 g of water at \(90^{\circ} \mathrm{C}\) added to a 500.0 g sample of water at \(10^{\circ} \mathrm{C}\). We can use a similar equation as in step 2, but now the masses are different: \(m_1c(T_f - T_{1}) = m_2c(T_{2} - T_f)\) Since the specific heat is equal, we can simplify the equation to: \(\frac{m_1}{m_2} = \frac{T_2 - T_f}{T_f - T_1}\) Plugging in the given values and solving for \(T_f\), we get: \(\frac{100}{500} = \frac{10 - T_f}{T_f - 90}\) Solving the equation, we find the final temperature: \(T_f \approx 22^{\circ} \mathrm{C}\) Thus, for situation b, the final temperature of the water should be between \(10^{\circ} \mathrm{C}\) and \(50^{\circ} \mathrm{C}\) (iii). ##Step 4: Applying the concept to situation c##

In situation c, we have 50.0 g of water at \(10^{\circ} \mathrm{C}\) with a 50.0 g iron ball at \(90^{\circ} \mathrm{C}\) being added. Since we have a different substance, we need to use the specific heat of iron, \(c_{Fe}\), instead of water. The equation for this situation is: \(m_{H_2O}c_{H_2O}(T_f - T_{H_2O}) = m_{Fe}c_{Fe}(T_{Fe} - T_f)\) Plugging in the given values and the known specific heats (for water, \(c_{H_2O} = 4.18\, J/g^{\circ}C\), and for iron, \(c_{Fe} = 0.45\, J/g^{\circ}C)\), we can solve for the final temperature: \(50 \times 4.18 (T_f - 10) = 50 \times 0.45 (90 - T_f)\) Solving this equation, we obtain: \(T_f \approx 25^{\circ} \mathrm{C}\) Thus, for situation c, the final temperature of the water should be between \(10^{\circ} \mathrm{C}\) and \(50^{\circ} \mathrm{C}\) (iii).

Key concepts

Specific Heat Capacity

Specific heat capacity measures how much energy a substance can hold per unit of mass. It’s like the individual's tolerance to heat. Different materials need different amounts of heat to raise their temperature by one degree Celsius.
For water, this value is quite high at 4.18 J/g°C, meaning water can absorb a lot of heat with little change in temperature. Iron, on the other hand, has a much lower specific heat capacity of 0.45 J/g°C. This means that iron heats up and cools down faster than water when exposed to the same amount of heat.

• The specific heat capacity of a substance is crucial in determining how much energy is exchanged in heat transfer processes.
• This factor plays a significant role when solving problems related to heat transfer, such as determining the final temperature when two substances are mixed or when one substance is heated.

Understanding this concept helps explain why mixing substances of different specific heat capacities results in varying temperature changes.

Thermal Equilibrium

Thermal equilibrium occurs when two substances reach the same temperature and heat transfer between them stops. It is a condition of "temperature balance" where an equal amount of heat is exchanged between the substances.
In the context of our problems, when two water samples, or a water sample and an iron ball, are mixed, they will exchange heat until they reach the same temperature. The concept tells us that the final temperature of a mixture is dictated by the ratio of masses and specific heat capacities of the substances involved.

• For instance, when equal masses of hot and cold water mix, they reach a midpoint temperature, as in situation a.
• However, when a small sample of hot water is mixed with a large cold water sample, the final temperature shifts significantly towards the colder temperature, as in situation b.

Understanding thermal equilibrium gives a clear insight into how temperatures behave when different thermal masses interact.

Conservation of Energy

The principle of conservation of energy is pivotal in understanding heat transfer. It states that energy cannot be created or destroyed, only transferred or converted from one form to another.
In heat transfer, all energy lost by a hot substance is gained by the cooler substance it is in contact with. This is the fundamental principle behind the equation used to solve the final temperatures in our problems.

• For example, when a hot iron ball is placed in cooler water, energy (heat) from the iron is transferred to the water until they reach thermal equilibrium.
• The total amount of heat lost by the iron equals the total gain in heat by the water, ensuring no energy is lost to the surroundings, assuming a perfectly insulated system.

A thorough understanding of energy conservation helps in setting up and solving heat equations, leading to the accurate prediction of final temperatures in thermodynamic problems.