A245.7g sample of metal at 75.0℃ was placed in 115.4g of water at 22.0℃. the final temperature of the water and metal was 34.0℃. if no heat was lost to the surroundings, what is the specific heat of the metal? (specific heat of water = 4.184 j/g℃)

The answer to your question is:  Cp = 0.6373 J/g°C

Explanation:

Sample

mass = 245.7 g

T1 = 75°C

T2 = 34°

Cp = ?

Water

Mass = 115.4 g

T1 = 22°C

T2 = 34°

Cp = 4.184 J/g°C

Heat of metal = mCp(T2 - T1) = 245.7Cp (34 - 75) = -9090.9Cp

Heat of water =                       = 115.4(4.184)(34 - 22) = -5794

                       Heat of metal = heat of water

                       -9090.9Cp = -5794

                       Cp = -5794 / -9090.9

                       Cp = 0.6373 J/g°C   specific heat of metal

The specific heat of the metal is

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

As we know that,  

        .................(1)

where,

q = heat absorbed or released

= mass of metal = 245.7 g

= mass of water = 115.4 g

= final temperature =

= temperature of metal =

= temperature of water =

= specific heat of metal = ?

= specific heat of water=

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

Therefore, the specific heat of the metal is

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the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be