When two samples of matter that are initially at different temperatures are placed in thermal contact, the samples undergo heat transfer until reaching thermal equilibrium at some new, final temperature. Transferred heat (q) can be measured based on the initial and final temperatures (Tinitial and Tfinal) of a sample that is placed into a known volume of water inside the insulated chamber of a calorimeter.Figure 1 Analysis of a heated sample using a calorimeterAccording to the first law of thermodynamics, the transferred heat energy is conserved, and the heat lost from the sample (indicated by a negative sign) must be equal to the sum of the heat gained by the water and the calorimeter, as expressed by Equation 1.−qsample = qwater + qcalorimeterEquation 1In a perfectly efficient system, qcalorimeter = 0, and all the heat released from the sample is retained by the water. In real systems, some heat is absorbed by the calorimeter itself. Based on the amount of heat transferred from the sample and the change in temperature, the heat capacity (C) of the entire sample can be determined. Expressing the heat capacity per unit mass yields the specific heat capacity (Cp) of the substance comprising the sample.Table 1 Measured specific heat capacities of several selected substancesSample Specific heatcapacity (J/g·°C)Lead 0.129Tungsten 0.132Silver 0.235Strontium 0.315Zinc 0.388Cobalt 0.421Titanium 0.525Wood 2.00Paraffin wax 2.5Water 4.184Experiment 1A sample of water (10.0 mL at 75 °C) was stirred into a calorimeter containing 100.0 mL of water initially at 21 °C. At thermal equilibrium, the system was found to have a temperature of 25 °C.Experiment 2The water in the calorimeter from Experiment 1 was discarded and replaced with 100.0 mL of fresh water at 25 °C. An unidentified 100.0 g sample of one of the metals from Table 1 was heated to an initial temperature of 80 °C and then placed into the water. At thermal equilibrium, the system was found to have a temperature of 30 °C. Question 59According to the results from Experiments 1 and 2, the unidentified metal sample is:A.Sr, because qcalorimeter > 0 and qmetal = qwater − qcalorimeterB.Co, because qcalorimeter ≈ 0 and qwater ≈ qmetalC.Ti, because qcalorimeter > 0 and qwater < qmetalD.unknown and its composition cannot be determined without more information.
Question
When two samples of matter that are initially at different temperatures are placed in thermal contact, the samples undergo heat transfer until reaching thermal equilibrium at some new, final temperature. Transferred heat (q) can be measured based on the initial and final temperatures (Tinitial and Tfinal) of a sample that is placed into a known volume of water inside the insulated chamber of a calorimeter.Figure 1 Analysis of a heated sample using a calorimeterAccording to the first law of thermodynamics, the transferred heat energy is conserved, and the heat lost from the sample (indicated by a negative sign) must be equal to the sum of the heat gained by the water and the calorimeter, as expressed by Equation 1.−qsample = qwater + qcalorimeterEquation 1In a perfectly efficient system, qcalorimeter = 0, and all the heat released from the sample is retained by the water. In real systems, some heat is absorbed by the calorimeter itself. Based on the amount of heat transferred from the sample and the change in temperature, the heat capacity (C) of the entire sample can be determined. Expressing the heat capacity per unit mass yields the specific heat capacity (Cp) of the substance comprising the sample.Table 1 Measured specific heat capacities of several selected substancesSample Specific heatcapacity (J/g·°C)Lead 0.129Tungsten 0.132Silver 0.235Strontium 0.315Zinc 0.388Cobalt 0.421Titanium 0.525Wood 2.00Paraffin wax 2.5Water 4.184Experiment 1A sample of water (10.0 mL at 75 °C) was stirred into a calorimeter containing 100.0 mL of water initially at 21 °C. At thermal equilibrium, the system was found to have a temperature of 25 °C.Experiment 2The water in the calorimeter from Experiment 1 was discarded and replaced with 100.0 mL of fresh water at 25 °C. An unidentified 100.0 g sample of one of the metals from Table 1 was heated to an initial temperature of 80 °C and then placed into the water. At thermal equilibrium, the system was found to have a temperature of 30 °C. Question 59According to the results from Experiments 1 and 2, the unidentified metal sample is:A.Sr, because qcalorimeter > 0 and qmetal = qwater − qcalorimeterB.Co, because qcalorimeter ≈ 0 and qwater ≈ qmetalC.Ti, because qcalorimeter > 0 and qwater < qmetalD.unknown and its composition cannot be determined without more information.
Solution
To answer this question, we need to calculate the specific heat capacity of the unknown metal from Experiment 2 and compare it with the values given in Table 1.
First, we need to calculate the heat gained by the water in the calorimeter. We can use the formula q = mcΔT, where m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
From the question, we know that the mass of the water is 100.0 g, the specific heat capacity of water is 4.184 J/g·°C, and the change in temperature is (30-25) °C = 5 °C.
So, qwater = (100.0 g) * (4.184 J/g·°C) * (5 °C) = 2092 J.
Next, we calculate the heat lost by the metal, which is equal to the heat gained by the water (since qmetal = -qwater). So, qmetal = -2092 J.
We can then calculate the specific heat capacity of the metal using the formula c = q / (mΔT), where m is the mass of the metal and ΔT is the change in temperature.
From the question, we know that the mass of the metal is 100.0 g and the change in temperature is (80-30) °C = 50 °C.
So, cmetal = -2092 J / [(100.0 g) * (50 °C)] = -0.4184 J/g·°C.
The negative sign indicates that the metal lost heat, which is expected. However, specific heat capacity is a positive quantity, so we take the absolute value: cmetal = 0.4184 J/g·°C.
Comparing this value with the values in Table 1, we see that it is closest to the specific heat capacity of Cobalt (Co), which is 0.421 J/g·°C.
Therefore, the unidentified metal sample is likely Cobalt (Co). So, the answer is B. Co, because qcalorimeter ≈ 0 and qwater ≈ qmetal.
Similar Questions
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