Question

Question #5b7cb

Answer 1

`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"`