# An object with a mass of 120 g is dropped into 640 mL of water at 0^@C. If the object cools by 48 ^@C and the water warms by 8 ^@C, what is the specific heat of the material that the object is made of?

Mar 21, 2016

c_(obj)=4.465 J/(g""^oC)

#### Explanation:

Given: An object with mass, ${m}_{o b j} = 120 g$
Object initial and final temperatures:
T_(obj)^i=48^oC; T_(obj)^f=8^oC
Water volume, ${V}_{{H}_{2} O} = 640 m L , \therefore {m}_{{H}_{2} O} = 640 g$
Water initial and final temperature:
T_w^i=0^oC; T_w^f=8^oC
Physics Principle of Heat: Heat Loss from the Object is equal to Heat gained by water, at equilibrium.
$\Delta {Q}_{l o s s} = {Q}_{\text{gained}}$
Heat Loss$\implies \Delta {Q}_{o b j} = {Q}_{f} - {Q}_{i} = m {c}_{o b j} \left({T}_{o b j}^{f} - {T}_{o b j}^{i}\right)$
Heat Gain$\implies \Delta {Q}_{w} = {Q}_{f} - {Q}_{i} = m {c}_{w} \left({T}_{w}^{f} - {T}_{w}^{i}\right)$
$m {c}_{o} \left({T}_{o b j}^{f} - {T}_{o b j}^{i}\right) = m {c}_{w} \left({T}_{w}^{f} - {T}_{w}^{i}\right)$ Substitute:
$120 \cdot \textcolor{red}{{c}_{o b j}} \left(48 - 8\right) = 640 \cdot 4.186 \left(8 - 0\right)$
Solve for $\textcolor{red}{{c}_{o b j}}$
c_(obj)=(64*cancel(8))/(12*cancel(40)^5)4.186=4.465 J/(g""^oC)