Question

17.) what is the specific heat of an unknown substance if 100.0 g of it at 200.0°c reaches an equilibrium temperature of 27.1°c when it comes in contact with a calorimeter of water? the water weighs 75 nd had an initial temperature of 20.0°c. (specific heat of water is 4.184 j/g°c).

Explanation:

Step1: Apply the principle of heat - transfer

The heat lost by the unknown substance is equal to the heat gained by the water. The formula for heat transfer is $Q = mc\Delta T$, where $Q$ is the heat transferred, $m$ is the mass, $c$ is the specific heat, and $\Delta T$ is the change in temperature.
Let $c_{u}$ be the specific heat of the unknown substance, $m_{u}=100.0\ g$ be its mass, $\Delta T_{u}=200.0 - 27.1=172.9^{\circ}C$ be its change in temperature, $m_{w} = 75\ g$ be the mass of water, $c_{w}=4.184\ J/g^{\circ}C$ be the specific - heat of water, and $\Delta T_{w}=27.1 - 20.0 = 7.1^{\circ}C$ be the change in temperature of water.
We have $Q_{u}=-Q_{w}$, so $m_{u}c_{u}\Delta T_{u}=m_{w}c_{w}\Delta T_{w}$.

Step2: Solve for the specific heat of the unknown substance

$c_{u}=\frac{m_{w}c_{w}\Delta T_{w}}{m_{u}\Delta T_{u}}$
Substitute the given values:
$m_{w} = 75\ g$, $c_{w}=4.184\ J/g^{\circ}C$, $\Delta T_{w}=7.1^{\circ}C$, $m_{u}=100.0\ g$, and $\Delta T_{u}=172.9^{\circ}C$ into the formula.
$c_{u}=\frac{75\times4.184\times7.1}{100.0\times172.9}$
First, calculate the numerator: $75\times4.184\times7.1 = 75\times29.7064=2227.98$.
Then, calculate the denominator: $100.0\times172.9 = 17290$.
$c_{u}=\frac{2227.98}{17290}\approx0.129\ J/g^{\circ}C$.

Answer:

$0.129\ J/g^{\circ}C$