Question

1. what mass of water can be heated from 25.0°c to 50.0°c by the addition of 2825 j? (the specific heat of water is 4.18 j/g°c.)

Explanation:

Step1: Calculate the temperature change

$\Delta T=T_{final}-T_{initial}=50.0 - 25.0=25.0^{\circ}C$

Step2: Use the heat - mass - specific heat formula

The formula for heat $Q = mc\Delta T$, where $Q$ is the heat energy, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the temperature change. We need to solve for $m$, so $m=\frac{Q}{c\Delta T}$.
Substitute $Q = 2825\ J$, $c = 4.18\ J/g^{\circ}C$, and $\Delta T=25.0^{\circ}C$ into the formula:
$m=\frac{2825}{4.18\times25.0}$
First, calculate the denominator: $4.18\times25.0 = 104.5$
Then, $m=\frac{2825}{104.5}=27.0\ g$

Answer:

$27.0\ g$