a 50 g sample of an unknown metal is heated to 90.0°c. it is placed in a perfectly insulated container along…

Explanation:

Step1: Determine heat lost by metal

The heat lost by the metal $Q_{metal}$ is given by the formula $Q = mc\Delta T$. Let the specific - heat capacity of the metal be $c_{metal}$. The mass of the metal $m_{metal}=50\ g$, the initial temperature of the metal $T_{i - metal}=90.0^{\circ}C$ and the final temperature $T_f = 25^{\circ}C$. So, $\Delta T_{metal}=T_f - T_{i - metal}=25 - 90=- 65^{\circ}C$. Then $Q_{metal}=m_{metal}c_{metal}\Delta T_{metal}=50\times c_{metal}\times(-65)=-3250c_{metal}$.

Step2: Determine heat gained by water

The heat gained by water $Q_{water}$ is given by $Q = mc\Delta T$. The mass of water $m_{water}=100\ g$, the specific - heat capacity of water $c_{water}=4.18\ J/g^{\circ}C$, the initial temperature of water $T_{i - water}=20^{\circ}C$ and the final temperature $T_f = 25^{\circ}C$. So, $\Delta T_{water}=T_f - T_{i - water}=25 - 20 = 5^{\circ}C$. Then $Q_{water}=m_{water}c_{water}\Delta T_{water}=100\times4.18\times5 = 2090\ J$.

Step3: Apply the principle of conservation of energy

In a perfectly insulated container, the heat lost by the metal is equal to the heat gained by the water (i.e., $Q_{metal}=-Q_{water}$). Since $Q_{metal}=-3250c_{metal}$ and $Q_{water}=2090\ J$, we have $-3250c_{metal}=-2090$.

Step4: Solve for the specific - heat capacity of the metal

Dividing both sides of the equation $-3250c_{metal}=-2090$ by $- 3250$, we get $c_{metal}=\frac{2090}{3250}\approx0.64\ J/g^{\circ}C$.

Answer:

$0.64$