Question

12. 175 g of Fe at 15.0oC is mixed with 225 g of H2O at 95.0oC. What is the final temperature of the mixture? The specific heats of iron= 0.448 J/g oC and water = 4.184 J/g oC). (keep in mind the following: heat gained + heat lost = 0 and q=mass x heat capacity x T)

Answer 1

Step 1,- Calculate the -

a / heat gained by iron = mass x specific heat x rise in temperature (or, = c x m x Delta t )

---------------------------------- = 175 x 0.448/ 4.184 x ( t_f  - 15.0 )

--------------------------------- = 175 x 0.1071 x ( t_f  - 15.0 )

---------------------------------- = 18.738 ( t_f  - 15.0 )

b / heat lost by water = 225 x 4.184/ 4.184 x (t_f  - 95.0 )

---------------------------------- = 225  x ( t_f   - 95.0 )

Note that,

(i) the specific heat is given as ( ----J / g o C ) , hence it should be converted into calories / g which represents smaller unit of heat , relating to smaller unit of weight ie . gm.

(ii) 1 calorie = 4.184 J   

2 / now apply the relation : heat gained + heat lost = 0 , & find out t_f  ( final temperature of the mixture in degrees Celcius . Thus,

18.738 ( t_f  - 15.0 ) + 225 ( t_f  -95.0 ) = 0

& t_f   = 88.85^oC

Answer - The final temperature of the mixture is 88.85^o C

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