Question

the specific heat capacity of fe is 0.449 j/g°c. a sample of fe was heated with 3.333 x 10³ j and the temperature was raised from 30.0 to 55.0 °c. (you dont need to convert j into cal). the mass of the fe sample is numerically 1 and the units are 2. a. 5.99 x 10¹ b. 1.67 x 10⁻² c. 1.86 x 10⁵ d. 5.39 x 10⁻⁶ e. 297 f. 3.37 x 10⁻³ g. 3.74 x 10⁴ h. 2.67 x 10⁻⁵ i. g j. °c k. j l. j/g °c m. g °c/j

Explanation:

Step1: Identify the formula

The formula for heat transfer is $q = mc\Delta T$, where $q$ is the heat energy, $m$ is the mass, $c$ is the specific - heat capacity, and $\Delta T$ is the change in temperature. We need to solve for $m$, so $m=\frac{q}{c\Delta T}$.

Step2: Calculate $\Delta T$

$\Delta T=T_2 - T_1=55.0^{\circ}C - 30.0^{\circ}C = 25.0^{\circ}C$.

Step3: Substitute values into the formula

Given $q = 3.333\times10^{3}\ J$, $c = 0.449\ J/g^{\circ}C$, and $\Delta T = 25.0^{\circ}C$. Then $m=\frac{3.333\times 10^{3}\ J}{0.449\ J/g^{\circ}C\times25.0^{\circ}C}$.
$m=\frac{3.333\times 10^{3}}{0.449\times25.0}\ g$.
$0.449\times25.0 = 11.225$.
$m=\frac{3.333\times 10^{3}}{11.225}\ g\approx297\ g$.

Answer:

1. E. 297
2. I. g