a 235 - gram sample of platinum released 1,23...
a 235 - gram sample of platinum released 1,230 joules of heat when it cooled to 74°c. what was the initial 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:
114°C
# Explanation:
## Step1: Recall the heat - transfer formula
$q = mc\Delta T$
where $q$ is the heat transferred, $m$ is the mass, $c$ is the specific heat, and $\Delta T$ is the change in temperature. Here, $q=- 1230\ J$ (negative because heat is released), $m = 235\ g$, and $c=0.133\ J/g^{\circ}C$.
## Step2: Calculate the change in temperature
$\Delta T=\frac{q}{mc}$
Substitute the values: $\Delta T=\frac{-1230\ J}{235\ g\times0.133\ J/g^{\circ}C}=\frac{-1230}{235\times0.133}\ ^{\circ}C\approx - 39.7^{\circ}C$
## Step3: Find the initial temperature
We know that $\Delta T=T_f - T_i$. Given $T_f = 74^{\circ}C$ and $\Delta T=-39.7^{\circ}C$.
So, $T_i=T_f-\Delta T$.
$T_i = 74^{\circ}C-(-39.7^{\circ}C)=74^{\circ}C + 39.7^{\circ}C=113.7^{\circ}C\approx114^{\circ}C$ (rounded to 3 significant figures)