Question #0c7c3

1 Answer
Aug 20, 2015

Answer:

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.