# An object with a mass of 20 g is dropped into 750 mL of water at 0^@C. If the object cools by 90 ^@C and the water warms by 18 ^@C, what is the specific heat of the material that the object is made of?

Mar 4, 2018

suppose,the specific heat of the object is $s C {g}^{-} {1}^{\circ} {C}^{-} 1$

So,during this process of achieving thermal equilibrium,the heat energy released by the object was taken by the water.

So,we can write,

$20 \cdot s \cdot 90 = 750 \cdot 1 \cdot 18$ (using, $H = m s d \theta$,where, $d \theta$ is the change in temperature,and specific heat for water is $1$ CGS unit and weight of $750 m L$ of water is $750 g$)

So,we get, $s = 7.5 C {g}^{-} {1}^{\circ} {C}^{-} 1$

Mar 4, 2018

The specific heat is $= 31.4 k J k {g}^{-} 1 {K}^{-} 1$

#### Explanation:

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

For the cold water,  Delta T_w=18ºC

For the hot object DeltaT_o=90ºC

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

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 ${m}_{o} = 0.020 k g$

The mass of the water is ${m}_{w} = 0.75 k g$

$0.020 \cdot {C}_{o} \cdot 90 = 0.75 \cdot 4.186 \cdot 18$

${C}_{0} = \frac{0.75 \cdot 4.186 \cdot 18}{0.020 \cdot 90}$

$= 31.4 k J k {g}^{-} 1 {K}^{-} 1$