# Question #0c7c3

##### 1 Answer

#### Answer:

You'd need **9400 J** of heat.

#### Explanation:

So, you're dealing with a 250-g sample of methanol at

Take a look at the value of the specific heat, which is given as **1 gram** of that substance by

In your case, if you supply **1 gram** of methanol with **2.51 J**, you increase its temperature by

Since you're dealing with more than one gram of methanol, and your aim is to increase its temperature by more than one degree Celsius, you can use this formula

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

One quick thing before applying the formula - when you're dealing with a difference between two temperatures, you have

#DeltaT_"Kelvin" = DeltaT_"Celsius"#

In your case, you have

#DeltaT_"Celsius" = 33 - 18 = 15^@"C"#

and

#DeltaT_"Kelvin" = (273.15 + 33) - (273.15 + 15)#

#DeltaT_"Kelvin" = color(red)(cancel(color(black)(273.15))) + 33 - color(red)(cancel(color(black)(273.15))) - 18 = "15 K"#

So, plug in your values and solve for

#q = 250color(red)(cancel(color(black)("g"))) * 2.51"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)("K")))) * (33 - 18)color(red)(cancel(color(black)("K")))#

#q = "9412.5 J" = color(green)("9400 J")#

The answer is rounded to two sig figs, the number of sig figs you gave for the mass of methanol used.