Question

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.

Answer 1

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.