An insulated container made of certain metal has a mass of 3.6 kg
and contains 14.0 kg of water. Both the container and the water
are initially at a temperature of 16°C. A 10.0 kg piece of the same
type of metal initially at a temperature of 50°C is dropped into the
water. After reaching thermal equilibrium, the entire system is at
18.0°C. What is the specific heat capacity of the metal? The
process takes place in a constant pressure.
To solve this problem, we can use the principle of conservation of energy. The heat gained by the water and container is equal to the heat lost by the metal.
First, let's calculate the initial thermal energy of the water and container:
Q_initial = mass_water * specific_heat_water * change_in_temperature_water + mass_container * specific_heat_container * change_in_temperature_container
The change in temperature for both the water and container is given as:
change_in_temperature_water = 18°C - 16°C = 2°C
change_in_temperature_container = 18°C - 16°C = 2°C
Now, let's calculate the heat gained by the water and container:
Q_water_container = 14.0 kg * specific_heat_water * 2°C + 3.6 kg * specific_heat_container * 2°C
Next, let's calculate the heat lost by the metal:
Q_metal = mass_metal * specific_heat_metal * change_in_temperature_metal
The change in temperature for the metal is given as:
change_in_temperature_metal = 18°C - 50°C = -32°C
Now, let's substitute the known values and solve for the specific heat capacity of the metal:
Q_water_container = Q_metal
14.0 kg * specific_heat_water * 2°C + 3.6 kg * specific_heat_container * 2°C = 10.0 kg * specific_heat_metal * (-32°C)
Simplifying the equation:
28.0 kg * specific_heat_water + 7.2 kg * specific_heat_container = -320 kg * specific_heat_metal
Divide the equation by the mass of the metal:
28.0 kg * specific_heat_water / 10.0 kg + 7.2 kg * specific_heat_container / 10.0 kg = -320 kg * specific_heat_metal / 10.0 kg
2.8 * specific_heat_water + 0.72 * specific_heat_container = -32 * specific_heat_metal
Now, we know that the specific heat capacity of the water is 4186 J/kg°C (assuming it's water at room temperature), and the specific heat capacity of the container is given as a certain value (which is not provided in the question). Let's call it C_container.
2.8 * 4186 J/kg°C + 0.72 * C_container = -32 * specific_heat_metal
Finally, we can solve for the specific heat capacity of the metal:
specific_heat_metal = (2.8 * 4186 J/kg°C + 0.72 * C_container) / (-32 kg)
The specific heat capacity of the metal depends on the specific heat capacity of the container, which is not provided in the question. Without that value, it is not possible to calculate the specific heat capacity of the metal.