Question

1. what is the specific heat capacity of silver metal if 55.00 g of the metal absorbs 47.3j of heat and the temperature rises 15.0°c?

Explanation:

Step1: Recall the heat - capacity formula

The formula for heat $Q = mc\Delta T$, where $Q$ is the heat absorbed or released, $m$ is the mass of the substance, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. We need to solve for $c$.
$c=\frac{Q}{m\Delta T}$

Step2: Substitute the given values

We are given that $Q = 47.3\ J$, $m=55.00\ g$, and $\Delta T = 15.0^{\circ}C$.
$c=\frac{47.3\ J}{55.00\ g\times15.0^{\circ}C}$

Step3: Calculate the specific - heat capacity

$c=\frac{47.3\ J}{825\ g\cdot^{\circ}C}\approx0.0573\ J/(g\cdot^{\circ}C)$

Answer:

$0.0573\ J/(g\cdot^{\circ}C)$