SOLUTION: A 50.0 g silver spoon at 20.0 0C is placed in a cup of coffee at 90.0 0C. How much heat will the cooler spoon absorb from the warmer coffee in order to reach a warmer temperature o

How much heat = 50*0.0558*(89-20) = 192.51 calories,    OR


              = 50*0.233*(89-20) = 803.85 joules.


Here 0.0558 = 0.0558 cal/(g*C) is the specific heat for silver,   also equal to

            = 0.233 J/(g*C)   in other dimension unit  (Joules instead of calories).



The general formula for absorbed heat is  H = M*c*dT,

where M is the mass, c is the specific heat capacity and dT is the temperature difference.