Question

What is the specific heat capacity of an unknown metal if 55.00 g of the metal initially at 90°C is placed in 75mL of water with an initial temperature of 25°C and the final temperature of the system is 35°C?

Answer 1

Answer:

The specific heat capacity of the unknown metal is 1.04 J/g°C.

Explanation:

Let the specific heat capacity of the unknown metal be $x$.

Given:

Mass of the metal is, $m=55\ g$

Initial temperature of the metal is, $T_i=90\ \°C$

Volume of water is, $V=75\ ml$

Specific heat capacity of water is, $c_w=4.186\ J/g\°C$

Initial temperature of water is, $T_{wi}=25\ \°C$

Final temperature of the system is, $T=35\ \°C$

We know that density of water is equal to 1 g/ml.

Mass is given as the product of density and volume.

Therefore, mass of water is given as:

$m_w=1\times 75=75\ g$

Now, fall of temperature of the unknown metal is given as:

$\Delta T_m=T_i-T=90-35=55\ \°C$

Rise of temperature of water is given as:

$\Delta T_w=T-T_{wi}=35-25=10\ \°C$

Now, as per conservation of energy,

Heat lost by metal = Heat gained by water

⇒ $mx\Delta T_m=m_wc_w\Delta T_w$

Plug in all the given values and solve for $x$. This gives,

$55\times x\times 55=75\times 4.186\times 10\\3025x=3139.5\\x=\frac{3139.5}{3025}=1.04\ J/g\°C$

Therefore, the specific heat capacity of the unknown metal is 1.04 J/g°C.