a 30.0 g sample of barium at 15.0 °c is warmed to 25.0 °c. how much energy is needed to warm the metal…

Explanation:

Step1: Identify the formula

The formula for heat energy is $q = mc\Delta T$, where $q$ is heat energy, $m$ is mass, $c$ is specific - heat capacity, and $\Delta T$ is the change in temperature.

Step2: Calculate the change in temperature

$\Delta T=T_{final}-T_{initial}=25.0^{\circ}C - 15.0^{\circ}C=10.0^{\circ}C$

Step3: Substitute the values into the formula

Given $m = 30.0\ g$, $c = 0.201\frac{J}{g\cdot^{\circ}C}$, and $\Delta T = 10.0^{\circ}C$. Then $q=mc\Delta T=(30.0\ g)\times(0.201\frac{J}{g\cdot^{\circ}C})\times(10.0^{\circ}C)$

Step4: Calculate the value of $q$

$q=(30.0)\times(0.201)\times(10.0)J = 60.3J$

Answer:

$60.3$