Chapter 6: Problem 12

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

Expert verified

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).