An object with a mass of #40 g# is dropped into #350 mL# of water. If the object cools by #25 ^@C# and the water warms by #10 ^@C#, what is the specific heat of the material that the object is made of?

Question

1873 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-26T12:29:39

#3.5# Cal per gram per degrees Celsius

#m_(water)*C_(water)*DeltaT=m_(object)*C_(object)*DeltaT#

#350*1*10 = 40*C_(object)*25#

#C_(object) = 3500/(40*25)#

#C_(object) = 3.5# #(Cal)/(g*degrees C)#

Your answer:

#C_(object) = 3.5# #(Cal)/(g*degrees C)#