Chemistry: Asample of nickel is heated to 99.8 degrees c and placed in a coffee cup calorimeter containing 150.0g water at 23.5 degrees c. after the metal cools, the final temperature of metal andwater mixture is 25.0 degrees c. if the specific heat capacity of nickel is 0.444j/(degreescg), what mass of nickel was originally heated?

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Asample of nickel is heated to 99.8 degrees c - id9984020951, glendaward88351

stankyweezle

11.01.2022

28.3 grams

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c_1\times \Delta T=m\times c_2\times (T_{final}-T_{initial})$

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

where,

q = heat absorbed or released

m_1 = mass of water= 150.0 g

m_2 = mass of nickel = ?

$T_{final}$ = final temperature = 25^oC=(273+25)K=298K

T_1 = temperature of water = 23.5^oC=(273+23.5)K=296.5K

T_2 = temperature of nickel = 99.8^oC=273+99.8=372.8K

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

c_2 = specific heat of nickel= 0.444J/g^0C

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

$150.0\times 4.184\times (298-296.5)=-[m_2\times 0.444\times (298-372.8)]$

m_2=28.3g

Therefore, the mass of nickel originally heated was 28.3 grams

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