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

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}}=(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* c_1* (T_f-T_1)=-m_2* c_2* (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* 0.89J/g^oC* (24.9-82.4)^oC=-m_2* 4.18J/g^oC* (24.9-22.3)^oC`

`m_2=200.12g`

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

Answer 2

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