[murdermaid]

If 450 cal of heat are added to 37 g of ethyl alcohol (s= 0.59 cal/g x C) at 20C, what would its final temperature be?

What I've tried doing is: q = 37g x 0.59 cal/g*C x 20C

I've tried adding 450 to the solution of this, doesn't seem to be the answer. If anyone can help it'd be greatly appreciated!

[coolman50544]

Specific heat can be defined as the amount of heat (measured in cal) for one unit of mass (measured in grams) required for something to heat up by one unit of temperature (measured in Celsius).

The specific heat you give tells us that 0.59 cal is needed for 1 gram of ethylene glycol to heat up by 1 degree Celsius.

We are considering 37 g of ethylene glycol. How much heat (in cal) is needed to heat up the whole 37 grams of ethylene glycol by one degree Celsius? For one gram, it is 0.59 cal. But what about for 37 grams? If you multiply the specific heat by 37 g, you will get this answer in units of cal/degree Celsius. I recommend that you search up something called dimensional analysis which is what is applied here.

You do not need to multiply by 20 degrees Celsius. Why do you think this is the case? Can you solve the rest of the problem by yourself?

[murdermaid]

This is my second week of class, and no I am not fluent in chemistry yet considering this an introductory course. Is this how the rest of the forum is going to act when I'm asking for instructions on how to build up an equation properly without googling something my professor hasn't even acknowledged yet?

[coolman50544]

This is my second week of class, and no I am not fluent in chemistry yet considering this an introductory course. Is this how the rest of the forum is going to act when I'm asking for instructions on how to build up an equation properly without googling something my professor hasn't even acknowledged yet?

Considering that dimensional analysis is something that may be taught as early as high school/middle school, yes.

[hypervalent_iodine]

If 450 cal of heat are added to 37 g of ethyl alcohol (s= 0.59 cal/g x C) at 20C, what would its final temperature be?

What I've tried doing is: q = 37g x 0.59 cal/g*C x 20C

I've tried adding 450 to the solution of this, doesn't seem to be the answer. If anyone can help it'd be greatly appreciated!

The equation you are using is Q = mcΔT. The ΔT term means change in temperature, and does not refer to your initial temperature of 20°C. This is the term you need to solve for, since you have Q, which is the heat, you have mass and you have c (specific heat capacity).

[Babcock_Hall]

Hi murdermaid,

According to the Forum Rules (see red link above), you must show your attempt first, before we can help you (IMO you met that burden).  In addition, we do not typically give answers; instead, we guide people to get there as much on their own as possible.

FWIW, I think that you initially misunderstood what the 20 °C meant.

[sunkal]

Hi murdermaid,

As you have written, Q = m s ΔT.  And ΔT = T_final - T_initial.  You have been given in the problem Q, m, s and T_initial.  Notice that 20 ^oC is T_initial.  You are asked to solve for T_final.  Using the first equation solve initially for ΔT.  Then using the definition of ΔT, solve for T_final.  Does this help you and can you do it now?