# An object with a mass of 10 g is dropped into 240 mL of water at 0^@C. If the object cools by 120 ^@C and the water warms by 8 ^@C, what is the specific heat of the material that the object is made of?

Feb 28, 2018

Here,the heat energy liberated by the object will be taken up by the water as a result of which,both will come to a thermal equilibrium.

So,heat energy liberated by the object is $10 \cdot s \cdot 120$ Calorie (using $H = m s d \theta$,where, $m$ is the mass of the object, $s$ is its specific heat and $d \theta$ is the change in temperature)

Now,heat energy absorbed by water is $240 \cdot 1 \cdot 1 \cdot 8$ calorie (where, mass of water = volume* density, and $1 m l = 1 c {m}^{3}$ and in CGS, density of water is $1 \frac{g}{c {m}^{3}}$ and specific heat of water is $1 C {g}^{-} 1 \circ {C}^{-} 1$)

So,equating both we get,

$1200 s = 1920$

so, $s = 1.6 C {g}^{-} 1 \circ {C}^{-} 1$

Feb 28, 2018

The specific heat is $= 6.76 .7 k J k {g}^{-} 1 {K}^{-} 1$

#### Explanation:

The heat is transferred from the hot object to the cold water.

For the cold water,  Delta T_w=8ºC

For the object DeltaT_o=120ºC

${m}_{o} {C}_{o} \left(\Delta {T}_{o}\right) = {m}_{w} {C}_{w} \left(\Delta {T}_{w}\right)$

The specific heat of water ${C}_{w} = 4.186 k J k {g}^{-} 1 {K}^{-} 1$

Let, ${C}_{o}$ be the specific heat of the object

The mass of the object is ${m}_{o} = 0.010 k g$

The mass of the water is ${m}_{w} = 0.24 k g$

$0.010 \cdot {C}_{o} \cdot 120 = 0.24 \cdot 4.186 \cdot 8$

${C}_{o} = \frac{0.24 \cdot 4.186 \cdot 8}{0.010 \cdot 120}$

$= 6.7 k J k {g}^{-} 1 {K}^{-} 1$