Question

part a
a 25.3 g sample of silver initially at 22.5 °c absorbs 150.0 j of heat. what is its final temperature?
view available hints
47.5 °c
25.5 °c
23.4 °c

Explanation:

Step1: Recall the heat - capacity formula

The formula for heat transfer 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 $\Delta T=T_f - T_i$ (final temperature - initial temperature). The specific heat capacity of silver, $c = 0.235\ J/g^{\circ}C$. We know $q = 150.0\ J$, $m=25.3\ g$ and $T_i = 22.5^{\circ}C$. First, we can re - arrange the formula for $\Delta T$: $\Delta T=\frac{q}{mc}$.

Step2: Calculate the change in temperature

Substitute the values into the formula for $\Delta T$: $\Delta T=\frac{150.0\ J}{25.3\ g\times0.235\ J/g^{\circ}C}=\frac{150.0}{25.3\times0.235}\ ^{\circ}C$.
$25.3\times0.235 = 5.9455$, so $\Delta T=\frac{150.0}{5.9455}\ ^{\circ}C\approx25.2\ ^{\circ}C$.

Step3: Calculate the final temperature

Since $\Delta T=T_f - T_i$, then $T_f=\Delta T + T_i$.
$T_f=25.2^{\circ}C+22.5^{\circ}C = 47.7^{\circ}C\approx47.5^{\circ}C$.

Answer:

A. $47.5^{\circ}C$