a 250.0g mass of metal at 112.4 degrees celcius is dropped in a calorimeter (c=1.40j/g degrees celsius, m = 5.0g) containing 150.0g of water. The water and calorimeter are initially at 25.0 degrees celsius. At the end of the experiment, the temperature of the water, calorimeter, and the metal is 37 degrees celsius. What is the celsius of the metal?
To find the specific heat of the metal (c of the metal), we can use the principle of energy conservation. The heat lost by the metal will be equal to the heat gained by the water and the calorimeter.
The amount of heat gained or lost by an object can be calculated using the formula:
Q = mcΔT
where:
Q is the heat gained or lost
m is the mass of the object
c is the specific heat capacity of the object
ΔT is the change in temperature
First, let's calculate the heat lost by the metal:
Q_lost = m_metal * c_metal * ΔT_metal
where:
m_metal is the mass of the metal (250.0g)
c_metal is the specific heat capacity of the metal (which we need to find)
ΔT_metal is the change in temperature of the metal (37°C - 112.4°C)
Q_lost = 250.0g * c_metal * (37°C - 112.4°C)
Now, let's calculate the heat gained by the water and calorimeter:
Q_gained = (m_water + m_calorimeter) * c_water * ΔT_water
where:
m_water is the mass of the water (150.0g)
m_calorimeter is the mass of the calorimeter (5.0g)
c_water is the specific heat capacity of water (1.40 J/g°C)
ΔT_water is the change in temperature of the water and calorimeter (37°C - 25°C)
Q_gained = (150.0g + 5.0g) * 1.40 J/g°C * (37°C - 25°C)
According to the principle of energy conservation, the heat lost by the metal is equal to the heat gained by the water and calorimeter:
Q_lost = Q_gained
250.0g * c_metal * (37°C - 112.4°C) = (150.0g + 5.0g) * 1.40 J/g°C * (37°C - 25°C)
Now, let's solve for c_metal:
c_metal = [(150.0g + 5.0g) * 1.40 J/g°C * (37°C - 25°C)] / [250.0g * (37°C - 112.4°C)]
Evaluate the expression on the right-hand side of the equation to find the specific heat capacity of the metal.