Question

1. graphite has a specific heat of 0.709 j/(g*°c). if a 25 gram piece of graphite is cooled from 35 °c to 18 °c. how much energy was lost by the graphite?

Explanation:

Step1: Calculate the temperature change

The formula for temperature change $\Delta T$ is $\Delta T=T_{final}-T_{initial}$.
Given $T_{initial} = 35^{\circ}C$ and $T_{final}=18^{\circ}C$, so $\Delta T=18 - 35=- 17^{\circ}C$.

Step2: Use the heat - transfer formula

The heat - transfer formula is $Q = mc\Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the temperature change.
Given $m = 25g$, $c=0.709J/(g\cdot^{\circ}C)$, and $\Delta T=-17^{\circ}C$.
Substitute the values into the formula: $Q=(25g)\times(0.709J/(g\cdot^{\circ}C))\times(-17^{\circ}C)$.
First, calculate $25\times0.709 = 17.725$.
Then, $17.725\times(-17)=-301.325J$.

Answer:

The energy lost by the graphite is $301.325J$ (the negative sign in the formula indicates heat loss, and we take the magnitude for the amount of energy lost).