19. how much will the temperature of a cup (180 g) of coffee at 95 °c be reduced when a 45 g silver spoon (specific heat 0.24 j/g oc) at 25 °c is placed in the coffee and the two are allowed to reach the same temperature assume that the coffee has the same density and specific heat as water.

The equation you will want to use for this relationship is:

Q = m*c*Δt
Where m is the mass (in grams), c is the specific heat capacity, and Δt is the change in temperature. 

The specific heat of the spoon is given (0.24 J/gC)
The specific heat for coffee is: 4.184 J/gC (I looked this up)

Let "x" be the final temperature of the system. 

So, our equations will be:

Heat lost by coffee = 180g * 4.184 J/gC * (95-x)
Heat gained by spoon = 45g * 0.24 J/gC * (x-25)

Next, we put our equations together:
Heat lost by coffee = Heat gained by spoon

So,

180 * 4.184 * (95-x) = 45 * 0.24 * (x-25)

Solve for x.

Finally, subtract the value you get for x from 95 (95⁰-x). This will give you how many degrees Celsius the coffee drops by. 

I hope this helps.

It conforms what ever surface its in.