A piece of metal of mass 12 g at 104◦C is
placed in a calorimeter containing 45.4 g of
water at 22◦C. The final temperature of the
mixture is 71.5
◦C. What is the specific heat
capacity of the metal? Assume that there is
no energy lost to the surroundings.
Answer in units of J
g ·
◦ C
I plug in the numbers and solve using a calculator, this is how i did it.
12(71.5-104)+45(71.5-22)=1857.3 J/g C
I still get the answer wrong.
Any errors i am making?
Sorry for posting twice :(
of course you get the wrong answer. What happened to cm,cw in my equation. You are looking for cm. Do some algebra.
12*Cm*(Tf-Tim)+45.4*Cw*(Tf-Tiw)=0
Tf=71.5 Tim=104 Tiw=22
solve for cm
To find the specific heat capacity of the metal, we need to calculate the heat transfer between the metal and the water.
The equation that relates heat transfer (Q), mass (m), specific heat capacity (c), and temperature change (ΔT) is:
Q = mcΔT
In this case, the metal is initially at 104°C and the water is initially at 22°C. The final temperature of the mixture is 71.5°C.
First, let's calculate the heat transfer from the metal to the water:
Q_metal = m_metal * c_metal * (T_final - T_initial)
Where:
m_metal = 12 g (mass of the metal)
c_metal = specific heat capacity of the metal (what we're trying to find)
T_initial = 104°C (initial temperature of the metal)
T_final = 71.5°C (final temperature of the mixture)
Next, let's calculate the heat transfer from the water:
Q_water = m_water * c_water * (T_final - T_initial)
Where:
m_water = 45.4 g (mass of the water)
c_water = specific heat capacity of water (4.18 J/g°C)
T_initial = 22°C (initial temperature of the water)
T_final = 71.5°C (final temperature of the mixture)
Since the system is isolated and assuming no energy is lost to the surroundings, the heat transfer from the metal is equal to the heat transfer to the water:
Q_metal = Q_water
So we can set up the equation:
m_metal * c_metal * (T_final - T_initial) = m_water * c_water * (T_final - T_initial)
Now, we can plug in the values and solve for c_metal:
12 g * c_metal * (71.5°C - 104°C) = 45.4 g * 4.18 J/g°C * (71.5°C - 22°C)
Simplifying the equation:
12 g * c_metal * (-32.5°C) = 45.4 g * 4.18 J/g°C * 49.5°C
Dividing both sides by (-32.5°C):
c_metal = (45.4 g * 4.18 J/g°C * 49.5°C) / (12 g * -32.5°C)
Calculating the value:
c_metal ≈ -7.02 J/g°C
Since specific heat capacity cannot be negative, we need to check our calculations or initial assumptions. Please double-check the data given and calculations performed.
The sum of the heats gained is zero.
heatgainedmetal+heatgained water=0
12*Cm*(Tf-Tim)+45.4*Cw*(Tf-Tiw)=0
Tf=71.5 Tim=104 Tiw=22
solve for cm