A coffee-cup calorimeter of the type shown in Figure 5.18 contains $150.0 \mathrm{~g}\( of water at \)25.2^{\circ} \mathrm{C}\(. A \)200-\mathrm{g}$ block of silver metal is heated to \(100.5^{\circ} \mathrm{C}\) by putting it in a beaker of boiling water. The specific heat of \(\mathrm{Ag}(s)\) is $0.233 \mathrm{~J} /(\mathrm{g} \cdot \mathrm{K})\(. The \)\mathrm{Ag}$ is added to the calorimeter, and after some time the contents of the cup reach a constant temperature of \(30.2^{\circ} \mathrm{C} .(\mathbf{a})\) Determine the amount of heat, in J, lost by the silver block. (b) Determine the amount of heat gained by the water. The specific heat of water is $4.184 \mathrm{~J} /(\mathrm{g} \cdot \mathrm{K}) .(\mathbf{c})$ The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam \(^{\circ}\) cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by \(1 \mathrm{~K} .\) Calculate the heat capacity of the calorimeter in \(\mathrm{J} / \mathrm{K}\). (d) What would be the final temperature of the system if all the heat lost by the silver block were absorbed by the water in the calorimeter?

Step by step solution

01

Find the heat lost by the silver block

To find the heat lost by the silver block, we'll use the equation \(q = mcΔT\), where \(m = 200~g\), \(c = 0.233~J/(g\cdot K)\), and \(ΔT = T_{initial} - T_{final} = 100.5^{\circ}C - 30.2^{\circ}C\). \(q_{silver} = (200)(0.233)(100.5-30.2) = 200 \times 0.233 \times 70.3~J\) Calculating the value, we get: \(q_{silver} = 3261.18~J\)

02

Calculate the heat gained by the water

Now, we will calculate the heat gained by water using the same equation \(q = mcΔT\), where \(m = 150~g\), \(c = 4.184~J/(g\cdot K)\), and \(ΔT = T_{final} - T_{initial} = 30.2^{\circ}C - 25.2^{\circ}C\). \(q_{water} = (150)(4.184)(30.2 - 25.2) = 150 \times 4.184 \times 5~J\) Calculating the value, we get: \(q_{water} = 3138~J\)

03

Calculate the heat capacity of the calorimeter

To calculate the heat capacity of the calorimeter, we will use the difference between the heat lost by the silver block and the heat gained by the water. We will also divide this difference by the total temperature change that occurred during the experiment (\(ΔT\)). \(Heat \, capacity = \frac{q_{silver} - q_{water}}{ΔT} = \frac{3261.18 - 3138}{5}~J/K\) Calculating the value, we get: \(Heat \, capacity = 24.636~J/K\)

04

Calculate the final temperature if all the heat lost by silver were absorbed by water

Finally, we will find the final temperature if all the heat lost by the silver block were absorbed by the water in the calorimeter. Since the heat lost by the silver block would be equal to the heat gained by the water, we can write: \(q_{silver} = mcΔT_{final}\) Solving for the final temperature change (\(ΔT_{final}\)), we get: \(ΔT_{final} = \frac{q_{silver}}{mc_{water}} = \frac{3261.18~J}{(150~g)(4.184~J/(g\cdot K))}\) Calculating the value, we get: \(ΔT_{final} = 5.18~K\) Now, we will add this temperature change to the initial temperature of water to get the final temperature of the system: \(T_{final} = T_{initial} + ΔT_{final} = 25.2^{\circ}C + 5.18~K\) Calculating the value, we get: \(T_{final} = 30.38^{\circ}C\)

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