Heat Effects and Calorimetry 
1. A metal sample weighing 147.90 g and at a temperature of 99.5°C was placed in 49.73 g of water in a calo- rimeter at 23.0°C. At equilibrium the temperature of the water and metal was 41.8°C. What was At for the water? (At = tfinal initial) a. b. What was At for the metal? C. d. Calculate the specific heat of the metal, using Equation 3. e. °C How much heat flowed into the water? (Take the specific heat of the water to be 4.18 J/g °C.) What is the approximate molar mass of the metal? (Use Eq. 4.) °C joules joules/g °C g/mol
The correct answer and explanation is:
Let’s break down the problem and solve each part step by step. We have a metal sample and water in a calorimeter, and we’re trying to calculate various quantities related to heat transfer.
Given Data:
- Mass of the metal sample (mmetalm_{\text{metal}}) = 147.90 g
- Initial temperature of the metal (Tinitial, metalT_{\text{initial, metal}}) = 99.5°C
- Mass of the water (mwaterm_{\text{water}}) = 49.73 g
- Initial temperature of the water (Tinitial, waterT_{\text{initial, water}}) = 23.0°C
- Final equilibrium temperature (TfinalT_{\text{final}}) = 41.8°C
- Specific heat capacity of water (cwaterc_{\text{water}}) = 4.18 J/g°C
a. Change in temperature for the water (ΔTwater\Delta T_{\text{water}}):
The formula for temperature change (ΔT\Delta T) is: ΔT=Tfinal−Tinitial\Delta T = T_{\text{final}} – T_{\text{initial}}
Substituting the values for water: ΔTwater=41.8∘C−23.0∘C=18.8∘C\Delta T_{\text{water}} = 41.8^\circ C – 23.0^\circ C = 18.8^\circ C
So, the change in temperature for the water is ΔTwater=18.8∘C\Delta T_{\text{water}} = 18.8^\circ C.
b. Change in temperature for the metal (ΔTmetal\Delta T_{\text{metal}}):
Now for the metal, the temperature change is calculated similarly: ΔTmetal=Tfinal−Tinitial, metal\Delta T_{\text{metal}} = T_{\text{final}} – T_{\text{initial, metal}}
Substitute the values for the metal: ΔTmetal=41.8∘C−99.5∘C=−57.7∘C\Delta T_{\text{metal}} = 41.8^\circ C – 99.5^\circ C = -57.7^\circ C
So, the change in temperature for the metal is ΔTmetal=−57.7∘C\Delta T_{\text{metal}} = -57.7^\circ C.
c. Calculate the specific heat of the metal (cmetalc_{\text{metal}}):
To find the specific heat of the metal, we can use the principle of conservation of energy. The heat lost by the metal is equal to the heat gained by the water: qmetal=−qwaterq_{\text{metal}} = -q_{\text{water}}
The heat transfer qq is given by: q=m⋅c⋅ΔTq = m \cdot c \cdot \Delta T
For the water: qwater=mwater⋅cwater⋅ΔTwaterq_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}
Substituting the known values: qwater=49.73 g⋅4.18 J/g°C⋅18.8∘C=3932.6 Jq_{\text{water}} = 49.73 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot 18.8^\circ C = 3932.6 \, \text{J}
Since heat lost by the metal equals heat gained by the water: qmetal=3932.6 Jq_{\text{metal}} = 3932.6 \, \text{J}
Now, we solve for the specific heat of the metal: qmetal=mmetal⋅cmetal⋅ΔTmetalq_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}} 3932.6 J=147.90 g⋅cmetal⋅(−57.7∘C)3932.6 \, \text{J} = 147.90 \, \text{g} \cdot c_{\text{metal}} \cdot (-57.7^\circ C)
Solving for cmetalc_{\text{metal}}: cmetal=3932.6147.90⋅57.7≈0.469 J/g°Cc_{\text{metal}} = \frac{3932.6}{147.90 \cdot 57.7} \approx 0.469 \, \text{J/g°C}
So, the specific heat of the metal is approximately 0.469 J/g°C0.469 \, \text{J/g°C}.
d. Molar mass of the metal:
To estimate the molar mass of the metal, we use the specific heat formula and compare the value to known specific heats of common metals. Typically, metals with a specific heat close to 0.469 J/g°C include certain metals like iron. To calculate the molar mass, we would need additional data such as the atomic or molecular structure of the metal. However, based on the specific heat, we can roughly estimate the metal could be iron, which has a molar mass of approximately 55.85 g/mol.
Thus, the approximate molar mass of the metal is around 55.85 g/mol, assuming it is iron.
Conclusion:
- The change in temperature of the water is 18.8°C.
- The change in temperature of the metal is -57.7°C.
- The specific heat of the metal is 0.469 J/g°C.
- The approximate molar mass of the metal is 55.85 g/mol.
These calculations demonstrate how calorimetry can be used to determine the specific heat and other properties of materials based on the heat transfer between them.