Question

A sample of aluminum, which has a specific heat capacity of 0.897·J·g−1°C^−1 , is put into a calorimeter (see sketch at right) that contains 300.0g of water. The aluminum sample starts off at 94.5°C and the temperature of the water starts off at 21.0°C .When the temperature of the water stops changing it's 23.8°C .The pressure remains constant at 1atm .Calculate the mass of the aluminum sample.

Answer 1

Answer: The mass of aluminium sample is 55.4 gram

Step-by-step explanation:

`Q_(absorbed)=Q_(released)`

As we know that,

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

`m_1* c* (T_(final)-T_1)=-[m_2* c* (T_(final)-T_2)]`

where,

`m_1` = mass of aluminium = ?

`m_2` = mass of water = 300.0 g

`T_(final)` = final temperature =
`23.8^0C`

`T_1` = temperature of aluminium =
`94.5^oC`

`T_2` = temperature of water =
`21.0^oC`

`c_1` = specific heat of aluminium =
`0.897J/g^0C`

`c_2` = specific heat of water =
`4.184J/g^0C`

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

`-[m_1* c_1* (T_(final)-T_1)]=[m_2* c_2* (T_(final)-T_2)]`

`-[m_1* 0.897* (23.8-94.5)^0C]=[300.0g* 4.184* (23.8-21.0)]`

`m_1=55.4g`

Therefore, the mass of aluminium sample is 55.4 gram