Question

1. 25.0 g of mercury is heated from 25°c to 155°c, and absorbs 455 joules of heat in the process. calculate the specific heat capacity of mercury.

Explanation:

Step1: Identify the heat - transfer formula

The heat - transfer formula is $Q = mc\Delta T$, where $Q$ is the heat absorbed or released, $m$ is the mass of the substance, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature.

Step2: Calculate the change in temperature

$\Delta T=T_{f}-T_{i}$, where $T_{f} = 155^{\circ}C$ and $T_{i}=25^{\circ}C$. So, $\Delta T=155 - 25=130^{\circ}C$.

Step3: Rearrange the formula to solve for $c$

From $Q = mc\Delta T$, we can solve for $c$ as $c=\frac{Q}{m\Delta T}$.

Step4: Substitute the given values

Given $Q = 455\ J$, $m = 25.0\ g$, and $\Delta T = 130^{\circ}C$. Substitute these values into the formula: $c=\frac{455\ J}{25.0\ g\times130^{\circ}C}$.

Step5: Calculate the value of $c$

$c=\frac{455}{25\times130}=\frac{455}{3250}=0.14\ J/(g\cdot^{\circ}C)$

Answer:

$0.14\ J/(g\cdot^{\circ}C)$