# Question 0c7c3

Aug 20, 2015

You'd need 9400 J of heat.

#### Explanation:

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

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

In your case, if you supply 1 gram of methanol with 2.51 J, you increase its temperature by ${1}^{\circ} \text{C}$.

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

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$m$ - the mass of the sample;
$c$ - the specific heat of methanol;
$\Delta T$ - the change in temperature, defined as ${T}_{\text{final" - T_"initial}}$.

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

$\Delta {T}_{\text{Kelvin" = DeltaT_"Celsius}}$

$\Delta {T}_{\text{Celsius" = 33 - 18 = 15^@"C}}$

and

$\Delta {T}_{\text{Kelvin}} = \left(273.15 + 33\right) - \left(273.15 + 15\right)$

$\Delta {T}_{\text{Kelvin" = color(red)(cancel(color(black)(273.15))) + 33 - color(red)(cancel(color(black)(273.15))) - 18 = "15 K}}$

So, plug in your values and solve for $q$

$q = 250 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 2.51"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)("K")))) * (33 - 18)color(red)(cancel(color(black)("K}}}}$

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.