# Question f09ab

Dec 19, 2016

${38.5}^{\circ} C$, rounded to one decimal place.

#### 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

$\Delta Q = m s t$, or $\Delta Q = m L$
where $m , s \mathmr{and} t$ are the mass, specific heat and rise or gain in temperature of the object;

$L$ is the latent heat for the change of state and

$\Delta {Q}_{\text{lost"=Delta Q_"gained}}$

In the given problem heat is lost by steam and gained by water aluminum container system. Let final temperature of the mixture be ${T}^{\circ} C$. Using CGS units.

Heat gained by water kept in container to change from temperature ${10}^{\circ} \text{C}$ to temperature at ${T}^{\circ} C$. $s$ for water =1calgm^-1"^@C^-1

$\Delta {Q}_{\text{gained 1}} = m s t = 200 \times 1 \times \left(T - 10\right) = 200 \left(T - 10\right)$
Heat gained by aluminum container to change from temperature ${10}^{\circ} \text{C}$ to temperature at ${T}^{\circ} C$.
$s$ for aluminum =0.215 calgm^-1"^@C^-1

$\Delta {Q}_{\text{gained 2}} = m s t = 50 \times 0.215 \times \left(T - 10\right) = 10.75 \left(T - 10\right)$

Total Heat gained by water and container $= \Delta {Q}_{\text{gained 1"+DeltaQ_"gained 2}} = 200 \left(T - 10\right) + 10.75 \left(T - 10\right)$ $= 210.75 \left(T - 10\right)$ .....(1)

Now heat lost by steam to cool down from 100^@"C# to become water at ${T}^{\circ} \text{C}$ is given by sum of heat lost in becoming water at ${100}^{\circ} C$, change of state, plus heat lost to become water at ${T}^{\circ} C$.
$L$ for steam $= 540 c a l g {m}^{-} 1$

$\Delta {Q}_{\text{lost 1}} = m L = 10 \times 540 = 5400$
$\Delta {Q}_{\text{lost 2}} = m s \left(100 - T\right) = 10 \times 1 \times \left(100 - T\right) = 10 \left(100 - T\right)$
Total heat lost by steam$= 5400 + 10 \left(100 - T\right)$ ......(2)

Equating (1) and (2) and solving for the required quantity
$5400 + 10 \left(100 - T\right) = 210.75 \left(T - 10\right)$
$\implies 5400 + 1000 - 10 T = 210.75 T - 2107.5$
$\implies 220.75 T = 8507.5$
$\implies T = {38.5}^{\circ} C$, rounded to one decimal place.