Question

1. graphite has a specific heat of ( 0.709 , \text{j/(g}\times^{circ}\text{c)} ). if a 25 gram piece of graphite is cooled from ( 35^{circ}\text{c} ) to ( 18^{circ}\text{c} ), how much energy was lost by the graphite?

Explanation:

Step1: Calculate the temperature change

$$\Delta T=T_{final}-T_{initial}=18 - 35=- 17^{\circ}C$$

Step2: Use the heat - transfer formula \(Q = mc\Delta T\)

Given \(m = 25g\), \(c=0.709J/(g\cdot^{\circ}C)\), \(\Delta T=-17^{\circ}C\)
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 calculation just indicates heat loss).