A piece of metal of mass 18 g at 104 ◦C is
placed in a calorimeter containing 47.4 g of
water at 22◦C. The final temperature of the
mixture is 40.8
◦C. What is the specific heat
capacity of the metal? Assume that there is
no energy lost to the surroundings.
That equation I gave you for the Fe and H2O works all of the heat problems involving two substance at different temperatures. Plug in the numbers and solve for the only unknown in the equation.
thank you!
To find the specific heat capacity of the metal, we can use the principle of conservation of energy. The heat gained by the metal is equal to the heat lost by the water.
The equation for heat gained or lost is given by:
Q = mcΔT
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
In this case, the heat gained by the metal is equal to the heat lost by the water. Let's calculate these values:
For the metal:
m = 18 g
ΔT = (40.8°C - 104°C) = -63.2°C (Negative sign indicates a decrease in temperature since the metal is cooling down.)
For the water:
m = 47.4 g
ΔT = (40.8°C - 22°C) = 18.8°C
Using the equation Q = mcΔT, we can calculate the heat gained by the metal and the heat lost by the water:
Q_metal = (18 g)(c_metal)(-63.2°C)
Q_water = (47.4 g)(4.18 J/g°C)(18.8°C) (The specific heat capacity of water is approximately 4.18 J/g°C.)
Since the heat gained by the metal is equal to the heat lost by the water, we can set up the equation:
Q_metal = Q_water
(18 g)(c_metal)(-63.2°C) = (47.4 g)(4.18 J/g°C)(18.8°C)
Now, we can solve for c_metal, the specific heat capacity of the metal:
c_metal = [(47.4 g)(4.18 J/g°C)(18.8°C)] / [(18 g)(-63.2°C)]
c_metal = -0.17 J/g°C
Therefore, the specific heat capacity of the metal is approximately -0.17 J/g°C.