Question

A 143-g sample of mercury is at an initial temperature of 25 °C. If 1067 joules of heat are applied to the sample, what is the final temperature of the mercury? The specific heat capacity of mercury is 0.14 J/(g∙ °C).

Answer 1

Following is the - complete Answer -&- Explanation: for the given Question, in.....typed format....

Answer:

Final temperature of the ( sample ) mercury : T_f = 78.30 ^oC   

Explanation:

Following is the complete Explanation, for the above Answer, in.....typed format...

• Given:

1. ​​​​​​​ Mass of the sample, of mercury: m = 143.0 g ( grams )
2. Initial Temperature of the sample; mercury : T_i = 25.0 ^oC
3. Amount of Heat applied to the sample: Q = 1067 J ( Joules )
4. Specific Heat Capacity of mercury : C = 0.14 J / (g . ^oC )

• ​​​​​​​Step - 1:

​​​​​​​ Let the final temperature , of the sample , -- be : T_f^oC   

We know the following: Equation of Heat Transfer :

Q = m x C x T   ------------------------------------------Equation - 1

Where:

1. m = mass of the sample ( i.e. mercury )
2. C = specific heat capacity of the sample ; [=] J / ( g. ^oC )
3. T = Change in temperature , of the sample, = ( T_f - T_i ) = ( T_f - 25.0 ^oC ) , i.e. in the given case.
4. Q =  Amount of Heat transferred, in the process of heat transfer....

• ​​​​​​​Step - 2:

​​​​​​​Plugging in the values in Equation - 1, as above, we would get the followiong:

Q = m x C x T   ------------------------------------------Equation - 1

  ( 1067 J ) = ( 143.0 g ) x ( 0.14 J / ( g. ^oC ) ) x (  T_f- 25.0 ^oC )

  (  T_f- 25.0 ^oC ) =   ( 1067 J ) / [  ( 143.0 g ) x ( 0.14 J / ( g. ^oC ) ) ] =  53.296 53.30  ^oC

T_f = ( 53.30 + 25.0 ) =  78.30 ^oC   .....

Final Temperature : T_f = 78.30 ^oC   

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