a 77 - gram sample of platinum at an initial ...
a 77 - gram sample of platinum at an initial temperature of 111°c absorbed 23 joules of heat. what is the final temperature of the platinum? the specific heat of platinum is 0.133 j/g°c. round your answer to the correct number of significant figures.
Answer
# Answer:
113 °C
# Explanation:
## Step1: Recall the heat - transfer formula
$q = mc\Delta T$, where $q$ is heat, $m$ is mass, $c$ is specific heat, and $\Delta T$ is change in temperature.
## Step2: Rearrange the formula for $\Delta T$
$\Delta T=\frac{q}{mc}$. Given $q = 23\ J$, $m = 77\ g$, and $c=0.133\ J/g^{\circ}C$.
$\Delta T=\frac{23}{77\times0.133}=\frac{23}{10.241}\approx2.25^{\circ}C$.
## Step3: Calculate the final temperature
$T_f=T_i+\Delta T$. Given $T_i = 111^{\circ}C$.
$T_f=111 + 2.25=113.25^{\circ}C$. Rounding to 3 significant figures gives $113^{\circ}C$.