# Question e69fc

Jun 14, 2016

$117.39 g$, rounded to two decimal places

#### Explanation:

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

Also that the heat gained/lost is given by
$\Delta Q = m s t$ ....(1)
where $m , s \mathmr{and} t$ are the mass, specific heat and rise or gain in temperature of the object;
$\Delta {Q}_{\text{lost"=Delta Q_"gained}}$ .....(2)

In the given problem heat is lost by sample of copper and gained by water.

Heat gained by water to change its temperature from ${21.0}^{\circ} \text{C}$ to ${29.7}^{\circ} \text{C}$
$\Delta {Q}_{\text{gained}} = m s t$, Specific heat of water is 1calg^-1 "^@C^-1.
$\Delta {Q}_{\text{gained}} = 298 \times 1 \times \left(29.7 - 21.0\right)$
$\implies \Delta {Q}_{\text{gained}} = 2592.6 c a l$

Heat lost by the sample to change its temperature from $275.1 {.}^{\circ} \text{C}$ to ${29.7}^{\circ} \text{C}$, specific heat of copper=0.09calg^-1 "^@C^-1#. If ${m}_{s}$ is the mass of sample
$\Delta {Q}_{\text{lost}} = {m}_{s} \times 0.09 \times \left(275.1 - 29.7\right)$
$\implies \Delta {Q}_{\text{lost}} = 22.086 \times {m}_{s} c a l$

Using (2) we get

$22.086 \times {m}_{s} = 2592.6$
$\implies {m}_{s} = 117.39 g$, rounded to two decimal places