# An object with a mass of 18 g is dropped into 400 mL of water at 0^@C. If the object cools by 150 ^@C and the water warms by 12 ^@C, what is the specific heat of the material that the object is made of?

Dec 28, 2017

$= \frac{7.4311 J}{g \cdot C}$

#### Explanation:

1. Given the volume of water used as medium to cool down the object, its mass can be derived through the density formula; that is,
$\rho = \frac{m}{V}$
$m = \rho \times V$
$m = \frac{1 g}{\cancel{\left(m l\right)}} \times 400 \cancel{m l}$
$m = 400 g {H}_{2} O$
2. Find the heat absorbed (gained) by the water that increases its temperature by ${12}^{o}$. Knowing the $C {p}_{w} = \frac{4.18 J}{g \cdot C}$, heat gained can be calculated by:
$Q = m C p \Delta T$
$Q = \left(400 \cancel{g}\right) \left(\frac{4.18 J}{\cancel{\left(g \cdot C\right)}}\right) \left(12\right) \cancel{^ o C}$
$Q = 20 , 064 J = 20.064 k J$
3. Since heat lost=heat gained; therefore, the $C p$ of the material can be computed as;
$Q = m C p \Delta T$
$C {p}_{o} = \frac{Q}{m \Delta T}$
$C {p}_{o} = \frac{20 , 064 J}{\left(18 g\right) \left({150}^{o} C\right)}$
$C {p}_{o} = \frac{7.4311 J}{g \cdot C}$