# Question #5b7cb

Jan 25, 2017

${60}^{\circ} \text{C}$

#### 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

$\Delta Q = m s t$,
where $m , s \mathmr{and} t$ are the mass, specific heat and rise or gain in temperature of the object;
and
$\Delta {Q}_{\text{lost"=Delta Q_"gained}}$

Let $T$ be equilibrium temperature. Using CGS units, heat lost by the object
$\Delta {Q}_{\text{lost}} = m \times 0.5 \times \left(80 - T\right)$ ........(1)

Heat gained by water
$\Delta {Q}_{\text{gained}} = m \times 1 \times \left(T - 20\right)$ ......(2)

Equating (1) and (2) we get
$m \times 0.5 \times \left(80 - T\right) = m \times 1 \times \left(T - 20\right)$
$\implies 0.5 \left(80 - T\right) = \left(T - 20\right)$

Multiplying both sides by $2$ we get
$\left(80 - T\right) = \left(2 T - 40\right)$
$\implies 3 T = 80 + 40$
$\implies T = \frac{120}{3} = {40}^{\circ} \text{C}$