Question

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.

Answer 1

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 2

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

Explanation :

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

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

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

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of aluminum = $0.89J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of Al = 42.5 g

$m_2$ = mass of water = ?

$T_f$ = final temperature of water = $24.9^oC$

$T_1$ = initial temperature of Al = $82.4^oC$

$T_2$ = initial temperature of water = $22.3^oC$

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

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

$m_2=200.12g$

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