1. Consider a calorimeter into which 135g of water at 45.3 degrees Celsius is added. The initial temperature of the calorimeter was 23.8C and the equilibrium temperature of the water and the calorimeter is 39.6C. Assuming there is no heat loss to the surroundings, what is the heat capacity of the calorimeter in J/C?
Isn't qH2O + qcalorimeter = 0?
q H2O = mass x specific heat x delta T.
q calor = mass x specific heat x delta T.
note: mass x specific heat calorimeter = calorimeter constant.
q H2O = 135 x 4.184 x (39.5-45.3)
q calorimeter = constant x (23.8-39.6)
To find the heat capacity of the calorimeter, we need to use the equation:
Q = m * c * ΔT
Where:
Q is the heat transferred,
m is the mass of the substance (in this case, the water),
c is the specific heat capacity of the substance (in this case, the water),
ΔT is the change in temperature.
First, let's calculate the heat transferred to the water using the equation above. The mass of the water is given as 135g, and the specific heat capacity of water is approximately 4.18 J/g°C.
Q_water = m * c * ΔT_water
ΔT_water = final temperature - initial temperature
= 39.6°C - 45.3°C
= -5.7°C
Q_water = (135g) * (4.18 J/g°C) * (-5.7°C)
= -3381.99 J (note that the negative sign indicates heat was lost by the water)
To find the heat capacity of the calorimeter, we need to rearrange the equation:
Q = m * c * ΔT
We know the heat transferred to the water (Q_water), and the temperature change (ΔT_calorimeter) of the water and the calorimeter, but we still need to calculate the mass and specific heat capacity of the calorimeter.
Since there is no heat loss to the surroundings, the heat transferred to the water is equal to the heat gained by the calorimeter:
Q_water = Q_calorimeter
We can rewrite this equation as:
m_water * c_water * ΔT_water = m_calorimeter * c_calorimeter * ΔT_calorimeter
We know the mass of the water (135g), the temperature change of the water and calorimeter (ΔT_calorimeter = -5.7°C), and the specific heat capacity of water (4.18 J/g°C). We need to solve for m_calorimeter * c_calorimeter.
Plugging in the values:
(135g) * (4.18 J/g°C) * (-5.7°C) = m_calorimeter * c_calorimeter * (-5.7°C)
Simplifying:
(-3381.99 J) = m_calorimeter * c_calorimeter * (-5.7°C)
Divide both sides of the equation by (-5.7°C):
m_calorimeter * c_calorimeter = -3381.99 J / (-5.7°C)
m_calorimeter * c_calorimeter = 592.63 J/°C
Therefore, the heat capacity of the calorimeter is 592.63 J/°C.
Note: The negative sign on the heat capacity indicates that the calorimeter absorbed heat from the water.