Question

Question #0c7c3

Answer 1

You'd need 9400 J of heat.

Explanation:

So, you're dealing with a 250-g sample of methanol at `18^@"C"`, and you want to figure out how much heat you need to supply in order to increase the sample's temeprature to `33^@"C"`.

Take a look at the value of the specific heat, which is given as `2.51"J/K g"`. A substance's specific heat tells you how much heat is needed to raise the temperature of 1 gram of that substance by `1^@"C"`.

In your case, if you supply 1 gram of methanol with 2.51 J, you increase its temperature by `1^@"C"`.

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

`m` - the mass of the sample;
`c` - the specific heat of methanol;
`DeltaT` - the change in temperature, defined as `T_"final" - T_"initial"`.

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`

`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.