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MakcuM [25]
3 years ago
12

A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate

r at a temperature of 23.0°C. After stirring, the final temperature of both copper and water is 26.0°C. Assuming no heat losses, and that the specific heat capacity of water is 4.184 J/(g·°C), what is the molar heat capacity of aluminum, Cm(Al)?
Physics
1 answer:
vivado [14]3 years ago
5 0

Answer: The molar heat capacity of aluminum is 25.3J/mol^0C

Explanation:

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of water = 130.0 g

m_2 = mass of aluminiunm = 23.5 g

T_{final} = final temperature = 26.0^oC=(273+26)K=299K

T_1 = temperature of water = 23^oC=(273+23)K=296K

T_2 = temperature of aluminium = 100^oC=273+100=373K

c_1 = specific heat of water= 4.184J/g^0C

c_2 = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]

c_2=0.938J/g^0C

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity =0.938J/g^0C\times 27g/mol=25.3J/mol^0C

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