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A 42.5 g piece of aluminum (which has a molar heat capacity of 24.03 j/ocmol) is heated to 82.4oc and dropped into a calorimeter containing water (specific heat capacity of water is 4.18 j/goc) initially at 22.3oc. the final temperature of the water is 24.9oc. calculate the mass of water in the calorimeter. ignore significant figures for this problem.

Respuesta :

1. Use the equation q = nC∆T
n = mols of aluminum = 42.5g/ 6.98g/mol
C = molar heat capacity = 24.03 j/Cmol
∆T = change in T from what it was before placed in calorimeter to after =
24.9C-82.4C (because water final and metal will be same temp.)
plug in to calculate q of metal =
q = (42.5/6.98)*(24.03J)(24.9-82.4)
q = -2176.5498 J
qmetal = -q of water
plug in values for water
-(-2176.5498 J) = mC∆T (m = mass of water in grams)
m = q/C∆T
∆T=24.9-22.3 C = 2.6
2176.5498 J/(2.6x4.184) = m = 200.2714 g

Answer : The mass of water in the calorimeter is, 200.12 grams.

Explanation :

First we have to calculate the specific heat capacity of aluminum.

[tex]\text{Specific heat capacity of Al}=\frac{\text{Molar heat capacity of Al}}{\text{Molar mass of Al}}=\frac{24.03J/mol^oC}{27g/mol}=0.89J/g^oC[/tex]

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

[tex]q_1=-q_2[/tex]

[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]

where,

[tex]c_1[/tex] = specific heat of aluminum = [tex]0.89J/g^oC[/tex]

[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]

[tex]m_1[/tex] = mass of Al = 42.5 g

[tex]m_2[/tex] = mass of water = ?

[tex]T_f[/tex] = final temperature of water = [tex]24.9^oC[/tex]

[tex]T_1[/tex] = initial temperature of Al = [tex]82.4^oC[/tex]

[tex]T_2[/tex] = initial temperature of water = [tex]22.3^oC[/tex]

Now put all the given values in the above formula, we get

[tex]42.5g\times 0.89J/g^oC\times (24.9-82.4)^oC=-m_2\times 4.18J/g^oC\times (24.9-22.3)^oC[/tex]

[tex]m_2=200.12g[/tex]

Therefore, the mass of water in the calorimeter is 200.12 grams.

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