An object with a mass of #50 g# is dropped into #300 mL# of water at #0^@C#. If the object cools by #80 ^@C# and the water warms by #8 ^@C#, what is the specific heat of the material that the object is made of?
1 Answer
Dec 7, 2017
Explanation:
#DeltaE# = change in energy (#J# )#m# = mass of the object (#kg# )#c# = specific heat capacity of the object (#J# #kg^(-1)# #K^(-1)# )#Deltatheta# = change in temperature (#K# )
Assuming all heat is transferred we have: