a .300 kg piece of metal is heated to 88 degrees celsius and placed in a copper calorimeter with a mass .150 kg which contains .500kg which contains .500 L of water that are both initially at 12.6 degrees celcius. Over short period of time the mixture comes to an equilibrium temperature of 22.5 degrees celcius. What is the specific heat of the metal?
1250j/kg oC
Cp= 0.1554 cal/g C
= 650.3453 J/kg K
To find the specific heat of the metal, you can use the principle of conservation of heat energy. The heat gained by the water and copper calorimeter must be equal to the heat lost by the metal.
The heat gained by the water and copper calorimeter can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat gained (or lost)
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
Let's start by calculating the heat gained by the water and copper calorimeter.
Mass of water = 0.5 kg
Specific heat of water = 4186 J/kg°C (approximately)
Change in temperature of water = (22.5 - 12.6) °C
Q_water = 0.5 kg * 4186 J/kg°C * (22.5 - 12.6) °C
Next, calculate the heat gained by the copper calorimeter.
Mass of copper calorimeter = 0.15 kg
Specific heat of copper = 390 J/kg°C (approximately)
Change in temperature of copper = (22.5 - 12.6) °C
Q_copper = 0.15 kg * 390 J/kg°C * (22.5 - 12.6) °C
Now, since the heat gained by the water and copper calorimeter is equal to the heat lost by the metal, we can equate the two equations:
Q_water + Q_copper = m_metal * c_metal * ΔT
Substituting the given values:
Q_water + Q_copper = 0.300 kg * c_metal * (22.5 - 88) °C
Now, solve for c_metal:
0.5 kg * 4186 J/kg°C * (22.5 - 12.6) °C + 0.15 kg * 390 J/kg°C * (22.5 - 12.6) °C = 0.300 kg * c_metal * (22.5 - 88) °C
Simplifying the equation, you can calculate the specific heat of the metal.