An object with a mass of 125 g is dropped into 750 mL of water at 0^@C. If the object cools by 70 ^@C and the water warms by 3 ^@C, what is the specific heat of the material that the object is made of?

Mar 2, 2016

$0.257$ to 3 decimal places

Explanation:

Specific heat Capacity is a measurement of the amount of energy needed to change the temperature of 1 Kg of the material by ${1}^{o}$C.

The name given to this energy type is Joules and has the symbol $J$.

Known:
Specific heat of water is 1
$1 m l \text{ of water weighs } 1 g$

The relevant association is
$\textcolor{b l u e}{\text{Mass "xx" change in temperature "xx" specific heat capacity}}$

Let $x$ be the Specific Heat Capacity of the object

For the object we have: $\text{ } 125 g \times {70}^{o} C \times x$
For the water we have:$\text{ } 750 \times {3}^{o} C \times 1$

Giving:

$750 \times {3}^{o} C \times 1 \text{ " =" "J" "=" } 125 g \times {70}^{o} C \times x$

$\implies x = \frac{3 \times 750}{125 \times 70}$

$0.257$ to 3 decimal places