# An object with a mass of 12 g is dropped into 240 mL of water at 0^@C. If the object cools by 120 ^@C and the water warms by 6 ^@C, what is the specific heat of the material that the object is made of?

Dec 6, 2016

0.5Cal//gm"^@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
$\Delta Q = m s \Delta t$
where $m , s \mathmr{and} \Delta t$ are the mass, specific heat and rise or gain in temperature of the object
$\Delta {Q}_{\text{lost"=Delta Q_"gained}}$

In the given problem heat is lost by the object and gained by water at ${0}^{\circ} C$. Using CGS system of units, where $1 m L$ of water has a mass of $1 g m$

Heat gained by water
$\Delta {Q}_{\text{gained}} = m s \Delta t$
$\Delta {Q}_{\text{gained}} = 240 \times 1 \times 6 = 1440 C a l$ ......(1)

And heat lost by object whose specific heat is ${s}_{o}$ is given by
$\Delta {Q}_{\text{lost"=m_"object}} \cdot {s}_{o} \cdot \Delta t$
$\implies \Delta {Q}_{\text{lost}} = 12 \cdot {s}_{o} \cdot 120$
$\implies \Delta {Q}_{\text{lost}} = 1440 {s}_{o}$ ......(2)
Equating (1) and (2) and solving for the required quantity
$1440 = 1440 {s}_{o}$
$\implies {s}_{o} = \frac{1440}{1440} = 1 C a l / g {m}^{\circ} C$