A 459.1-g sample of an element at 187°C is dropped into an ice–water mixture; 102.5 g of ice melts and an ice–water mixture remains. Calculate the specific heat of the element. ΔHfusion = 6.02 kJ/mol (for liquid water at 0°C).
heat lost by metal is
mass metal x specific heat x (Tfinal-Tinitial). Tf = 0; Ti = 187
heat gained by ice is
mass ice x heat fusion. I used 333.5 J/g for heat fusion
Put those together.
(mass metal x specific heat metal x (Tf-Ti) - (mass ice x heat fusion ice) = 0
Solve for specific heat metal. Something like 0.4 or so I think. That's an estimate.
I added those together but put a - sign. Arrrgh!. The following corrects that.
(mass metal x specific heat metal x (Tf-Ti) + (mass ice x heat fusion ice) = 0
To calculate the specific heat of the element, we need to use the equation:
q = m * c * ΔT
Where:
q is the heat absorbed or released by the substance,
m is the mass of the substance,
c is the specific heat of the substance,
ΔT is the change in temperature.
First, let's calculate the heat released by the element as it cools down from 187°C to 0°C. Since you mentioned that an ice-water mixture remains, we can assume that the water/ice mixture is at 0°C.
q1 = m1 * c1 * ΔT1
Here:
m1 is the mass of the sample at 187°C, which is 459.1g.
c1 is the specific heat of the element, which we are trying to find.
ΔT1 is the change in temperature, which is 187°C - 0°C = 187°C.
To find q1, we need to know the specific heat of the element. We can calculate it using the equation:
q1 = ΔHfusion * n
Where:
q1 is the heat absorbed or released by the element during the phase change.
ΔHfusion is the enthalpy of fusion of water, which is 6.02 kJ/mol (given in the problem).
n is the number of moles of water melted.
To find n, we need to calculate the number of moles of water melted.
n = m2 / M2
Here:
m2 is the mass of ice melted, which is 102.5 g.
M2 is the molar mass of water, which is 18 g/mol.
Substituting the values:
n = 102.5 g / 18 g/mol
n = 5.6944 mol
Now that we know the value of n, we can calculate q1:
q1 = 6.02 kJ/mol * 5.6944 mol
q1 = 34.284 kJ
Next, we can calculate q1 using the first equation:
q1 = m1 * c1 * ΔT1
34.284 kJ = 459.1 g * c1 * 187°C
Now solve for c1:
c1 = 34.284 kJ / (459.1 g * 187°C)
c1 = 0.40 J/g°C
Therefore, the specific heat of the element is 0.40 J/g°C.