A 0.050 0-kg ingot of metal is heated to 200.0°C and then dropped into a beaker containing 0.400 kg of water initially at 20.0°C. If the final equilibrium temperature of the mixed system is 22.4°C, find the specific heat of the metal.
Question : A 0.050 0-kg ingot of metal is heated to 200.0°C and then dr...
- Mass of the ingot of metal: 0.050 kg
- Initial temperature of the ingot of metal: 200.0°C
- Mass of water in the beaker: 0.400 kg
- Initial temperature of the water: 20.0°C
- Final equilibrium temperature of the mixed system: 22.4°C
Step 1:
The given equation is derived from Equation 20.5, which relates the heat capacities of two substances in thermal contact. The equation is:
mωcω(Tf-Tω) = -m_xc_x(T_f-T_x)
where mω and cω are the mass and specific heat capacity of substance ω, Tf is the final equilibrium temperature, Tω is the initial temperature of substance ω, m_x and c_x are the mass and specific heat capacity of substance x, and T_x is the initial temperature of substance x.
In this case, we are given the values for mω, cω, Tf, Tω, m_x, and T_x, and we are asked to find the value of c_x.
Step 2:
To solve for c_x, we can rearrange the equation as follows:
cx = -(mωcω(T_f-Tω))/(m_x(T_f-T_x))
Step 3:
Substituting the given values into the equation, we can calculate the value of c_x as:
c_x = -(0.400 kg)(4186 J/kg°C)(22.4°C-20.0°C)/-(0.0500 kg)(22.4°C-200.0°C)
Simplifying the expression, we find that c_x = 459 J/kg°C.
Step 4:
Based on the comparison of this result with the data given in Table 20.1, we can conclude that the ingot is most likely made of iron. Additionally, we note that the initial temperature of the ingot is above the steam point, so some of the water may vaporize when the ingot is dropped into the water. However, assuming a sealed system, any steam that forms will recondense back into water due to the final equilibrium temperature being lower than the steam point.
Final Answer
According to Equation 20.5 , we can write
m_{\omega} c_{\omega}\left(T_{f}-T_{\omega}\right)=-m_{x} c_{x}\left(T_{f}-T_{x}\right) (0.400 \mathrm{~kg})\left(4186 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(22.4^{\circ} \mathrm{C}-20.0^{\circ} \mathrm{C}\right)= -(0.0500 \mathrm{~kg})\left(c_{x}\right)\left(22.4^{\circ} \mathrm{C}-200.0^{\circ} \mathrm{C}\right)From this we find that
c_{x}=459 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}The ingot is most likely iron, as we can see by comparing this result with the data given in Table 20.1. Note that the temperature of the ingot is initially above the steam point. Thus, some of the water may vaporize when we drop the in- got into the water. We assume that we have a sealed system and thus that this steam cannot escape. Because the final equilibrium temperature is lower than the steam point, any steam that does result recondenses back into water.
| TABLE 20.1 Specific Heats of Some Substances at 25°C and Atmospheric Pressure | ||
| Substance | \frac{\text { Specific Heat } c}{\mathrm{~J} / \mathrm{k} \mathrm{g} \cdot{ }^{\circ} \mathrm{C} \quad \text { cal } / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}} | |
| Elemental Solids | ||
| Aluminum | 900 | 0.215 |
| Beryllium | 1 830 | 0.436 |
| Cadmium | 230 | 0.055 |
| Copper | 387 | 0.092 4 |
| Germanium | 322 | 0.077 |
| Gold | 129 | 0.030 8 |
| Iron | 448 | 0.107 |
| Lead | 128 | 0.030 5 |
| Silicon | 703 | 0.168 |
| Silver | 234 | 0.056 |
| Other Solids | ||
| Brass | 380 | 0.092 |
| Glass | 837 | 0.2 |
| Ice ( 5°C) | 2 090 | 0.5 |
| Marble | 860 | 0.21 |
| Wood | 1 700 | 0.41 |
| Liquids | ||
| Alcohol (ethyl) | 2 400 | 0.58 |
| Mercury | 140 | 0.033 |
| Water (15°C) | 4 186 | 1 |
| Gas | ||
| Steam (100°C) | 2 010 | 0.48 |