# Question #a6d1a

##### 1 Answer

#### Explanation:

As you know, a substance's specific heat tells you how much heat is required to increase the temperature of

In your case, graphite is said to have a specific heat equal to **units** for the specific heat of graphite and the units *given to you* in the problem.

First thing to notice here is that you are given a mass of graphite expressed in *kilograms*, but that the specific heat values uses *grams*. So a unit conversion is needed here.

Likewise, the temperature increase is given to you in *Kelvin*, but the specific heat value uses *degrees Celsius*.

Now, when it comes to **increase** or **decrease** in temperature, it makes **no difference** which temperature scale is used. For example, the change in temperature will be

#DeltaT = "348 K" - "294 K" = "54 K"#

Calculated in degrees Celsius, you have

#DeltaT = (348 -273.15)^@"C" - (294 - 273.15)^@"C"#

#DeltaT = 348^@"C" - color(red)(cancel(color(black)(273.15^@"C"))) - 294^@"C" + color(red)(cancel(color(black)(273.15^@"C"))) = 54^@"C"#

This means that you can use the *change in temperature* interchangeably for Kelvin and degrees Celsius.

Now, the equation that establishes a relationship between heat absorbed/lost and change in temperature looks like this

#color(blue)(q = m * c * DeltaT)" "# , where

*final temperature* minus *initial temperature*

Plug in your values and solve for **do not** forget about converting the mass of graphite from *kilograms* to *grams*

#q = 63 color(red)(cancel(color(black)("kg"))) * (1000color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * 0.71"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * 54color(red)(cancel(color(black)(""^@"C")))#

#q = "2,415,420 J"#

Rounded to two sig figs, the answer will be