4. A student performs an experiment to determine the specific heat capacity of a metal. The student — Thermodynamics, Calorimetry, Particulate Diagrams Chemistry Question
Problem Context
- A student performs an experiment to determine the specific heat capacity of a metal. The student places a cube of the metal in boiling water so its temperature will be 100.0°C. The student then places the metal cube into a calorimeter that contains water and records the highest temperature of the water. A data table and a diagram of the thermometer at the highest temperature are shown.
(a) What should the student report as the highest temperature of the water? _____________
Model Answer
38.5 °C [1, 2].
(b) A particle-level representation of water molecules in the calorimeter before and after the metal cube was added is shown.
The length of the arrows in the Before diagram represents the speed of the water molecules in the system. In the After diagram, draw an arrow for each molecule to indicate how the speed of each of the molecules changes after the metal cube is added.
Model Answer
The “After” drawing should contain arrows that are longer, on average
[1, 2].
(c) Assuming the metal transfers 2940 J of thermal energy to the water, calculate the specific heat of the metal in J/(g·°C).
Model Answer
*q = mcΔT*
*c_metal* = *q_metal* / (*m_metal* × *ΔT_metal*) = -2940 J / ((98.1 g)(38.5 °C -
100.0 °C)) = 0.487 J/(g·°C) [1, 2].
(d) In a second experiment, 2940 J of thermal energy is transferred from 98.1 g of aluminum, which has a specific heat capacity of 0.897 J/(g·°C). Explain how the magnitude of the temperature change of the aluminum, ΔT_Al, compares with the magnitude of the temperature change of the metal in the original experiment.
Model Answer
Accept one of the following:
* The value of *ΔT_Al* will be smaller because Al has a greater specific heat
capacity than the metal in the original experiment [1, 2]. Therefore, the same
thermal energy transfer applied to the same mass will result in a smaller change
in temperature, according to the equation *q = mcΔT* [1, 2].
* *q = mcΔT*
|*ΔT_Al*| = |*q_Al* / (*m_Al* × *c_Al*)| = | -2940 J / ((98.1 g)(0.897
J/(g·°C))) | = 33.4 °C [1, 2].
|*ΔT_metal*| = | 38.5 °C - 100.0 °C | = 61.5 °C [1, 2].
Thus, *ΔT_Al* < *ΔT_metal* [1, 2].
New conversation: e58311e8-5544-453d-aa46-7d2a30dcb458