A calorimeter is filled with 250.0 g of water at 24.6oC. A 60.0 g sample of a metal at 100.0 oC is dropped in this calorimeter and causes the temperature to increase a total of 4.5oC. What is the specific heat of the metal (in J/g°C)? Ignore the calorimeter in this problem
heat lost by metal is gained by water
water mass * specific heat * temp change = metal mass * specific heat * temp change
To find the specific heat of the metal, we can use the equation:
q = m * c * ΔT
Where:
q is the heat transferred
m is the mass of the substance
c is the specific heat capacity
ΔT is the change in temperature
In this case, the heat transferred (q) can be calculated by finding the heat gained by the water and the heat lost by the metal, assuming no heat is lost to the surroundings.
1. Calculate the heat gained by the water:
The mass of water is given as 250.0 g. The specific heat capacity of water is approximately 4.18 J/g°C. The initial temperature of water is 24.6°C, and it increases by 4.5°C.
q_water = m_water * c_water * ΔT_water
q_water = 250.0 g * 4.18 J/g°C * 4.5°C
2. Calculate the heat lost by the metal:
The mass of the metal is given as 60.0 g. Assuming the density of the metal is uniform, we can consider the entire mass of the metal as one temperature. The initial temperature of the metal is 100.0°C, and it decreases by 4.5°C.
q_metal = m_metal * c_metal * ΔT_metal
q_metal = 60.0 g * c_metal * (-4.5°C)
3. Set the heat gained by the water equal to the heat lost by the metal, since they both contribute to the change in temperature of the system:
q_water = q_metal
250.0 g * 4.18 J/g°C * 4.5°C = 60.0 g * c_metal * (-4.5°C)
Simplify the equation by canceling out the units:
(250.0 g * 4.18 J/g°C * 4.5°C) = (60.0 g * c_metal * (-4.5°C))
Solve for c_metal by isolating it:
c_metal = (-250.0 g * 4.18 J/g°C * 4.5°C) / (60.0 g * (-4.5°C))
4. Calculate the value of c_metal using the given values:
c_metal = (-46,725 J°C) / (-270.0 g°C)
c_metal ≈ 173.4 J/g°C
Therefore, the specific heat of the metal is approximately 173.4 J/g°C.