The specific heat of a certain type of metal is 0.128 J/(g·°C). What is the final temperature if 305 J of heat is added to 83.0 g of this metal initially at 20.0 °C?

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Answer 1 · 2016-11-17T17:12:10

We know the heat gained

#DeltaH=mxxsxx(t_2-t_1)#

where

#DeltaH=305J#
#m->"mass"=83g#

#s->"sp heat"=0.128" J/(g^@C)#

#t_1->"initial temperature"=20^@C#

#t_2->"final temperature"=?#

So

#DeltaH=mxxsxx(t_2-t_1)#

#=>305J=83gxx0.128J/(g^@C)xx(t_2-20)^@C#

#=>t_2-20=305/(83xx0.128)#

#=>t_2=48.7^@C#