Question

a 50 g sample of an unknown metal is heated to 90.0°c. it is placed in a perfectly insulated container along with 100 g of water at an initial temperature of 20°c. after a short time, the temperature of both the metal and water become equal at 25°c. the specific heat of water is 4.18 j/g°c in this temperature range. what is the specific heat capacity of the metal? record your answer with two significant figures. j/g°c

Explanation:

Step1: Determine heat gained by water

The heat - transfer formula is $Q = mc\Delta T$. For water, $m_{w}=100\ g$, $c_{w}=4.18\ J/g^{\circ}C$, $\Delta T_{w}=25 - 20=5^{\circ}C$. So $Q_{w}=m_{w}c_{w}\Delta T_{w}=100\times4.18\times5 = 2090\ J$.

Step2: Determine heat lost by metal

In an insulated system, the heat lost by the metal ($Q_{m}$) is equal to the heat gained by the water ($Q_{w}$), so $Q_{m}=- 2090\ J$ (negative because heat is lost). The mass of the metal $m_{m}=50\ g$, and $\Delta T_{m}=25 - 90=-65^{\circ}C$.

Step3: Calculate specific heat of metal

Using the formula $Q = mc\Delta T$ for the metal, we can solve for $c_{m}$. Rearranging gives $c_{m}=\frac{Q_{m}}{m_{m}\Delta T_{m}}$. Substituting the values: $c_{m}=\frac{-2090}{50\times(-65)}\approx0.64\ J/g^{\circ}C$.

Answer:

$0.64$