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

take a better picture and ill do it

a is the answer to this question

explanation: