it requires 350 joules to raise a certain amount of a substance from 10.0°c to 30.0°c. the specific heat of…

Explanation:

Step1: Identify the heat - transfer formula

The heat - transfer formula is $Q = mc\Delta T$, where $Q$ is the heat energy transferred, $m$ is the mass of the substance, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature.

Step2: Calculate the change in temperature

$\Delta T=T_2 - T_1=30.0^{\circ}C - 10.0^{\circ}C = 20.0^{\circ}C$

Step3: Rearrange the formula to solve for mass

From $Q = mc\Delta T$, we can solve for $m$: $m=\frac{Q}{c\Delta T}$. We know that $Q = 350\ J$, $c = 1.2\ J/g^{\circ}C$, and $\Delta T=20.0^{\circ}C$. Substitute the values into the formula: $m=\frac{350\ J}{1.2\ J/g^{\circ}C\times20.0^{\circ}C}=\frac{350}{24}\ g\approx14.58\ g\approx15\ g$

Answer:

15 g