Question #5b7cb

1 Answer
Jan 25, 2017

Answer:

#60^@"C"#

Explanation:

We know that in such interactions heat is gained by one and heat is lost is by the other.

Also that the heat gained/lost is given by

#DeltaQ=mst#,
where #m,s and t# are the mass, specific heat and rise or gain in temperature of the object;
and
#Delta Q_"lost"=Delta Q_"gained"#

Let #T# be equilibrium temperature. Using CGS units, heat lost by the object
#Delta Q_"lost"=mxx0.5xx(80-T)# ........(1)

Heat gained by water
#Delta Q_"gained"=mxx1xx(T-20)# ......(2)

Equating (1) and (2) we get
#mxx0.5xx(80-T)=mxx1xx(T-20)#
#=>0.5(80-T)=(T-20)#

Multiplying both sides by #2# we get
#(80-T)=(2T-40)#
#=>3T=80+40#
#=>T=120/3=40^@"C"#