Question

if 801 j of heat is available, what is the mass in grams of iron (specific heat = 0.450 j/g·°c) that can be heated from 22.5 °c to 120.0 °c?
answer:
g

Explanation:

Step1: Calculate the temperature change

The formula for temperature change is $\Delta T=T_{final}-T_{initial}$.
Substituting the given values: $\Delta T = 120.0 - 22.5=97.5^{\circ}C$.

Step2: Use the heat - mass formula

The formula for heat is $Q = mc\Delta T$, where $Q$ is heat, $m$ is mass, $c$ is specific heat.
We need to solve for $m$, so $m=\frac{Q}{c\Delta T}$.
Substituting $Q = 801\ J$, $c=0.450\ J/g\cdot^{\circ}C$, and $\Delta T = 97.5^{\circ}C$:
$m=\frac{801}{0.450\times97.5}$.
First, calculate $0.450\times97.5 = 43.875$.
Then, $m=\frac{801}{43.875}=18.26\ g$.

Answer:

$18.3\ g$ (rounded to three significant figures)