# An object with a mass of 15 g is dropped into 250 mL of water at 0^@C. If the object cools by 120 ^@C and the water warms by 9 ^@C, what is the specific heat of the material that the object is made of?

Apr 30, 2018

From your data we can assume the object lost heat and the water gained heat in equilibrating.

Moreover, I am assuming:

rho_(H_2O) = (1.0"g")/"mL"

Recall,

$q = m {C}_{s} \Delta T$

In this case, the heat gained by the water is supplied by the body. Accordingly, we can rearrange the equation, such that,

$- {m}_{x} {C}_{x} \Delta T = {m}_{{H}_{2} O} {C}_{{H}_{2} O} \Delta {T}_{{H}_{2} O}$

Hence,

$\implies {C}_{x} = \frac{- {m}_{{H}_{2} O} {C}_{{H}_{2} O} \Delta {T}_{{H}_{2} O}}{m x \Delta {T}_{x}} \approx \left(5.23 \text{J")/("g"*°"C}\right)$

is the specific heat of the body of mass we dropped into the water.