Question #5b7cb

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Answer 1 · 2017-01-25T00:07:03

$60^@"C"$

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
$(80-T)=(2T-40)$
$=>3T=80+40$
$=>T=120/3=40^@"C"$