Chemical Forums

Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: stag on March 24, 2011, 02:41:24 PM

Title: Heat lost and Heat gained
Post by: stag on March 24, 2011, 02:41:24 PM: I have a couple questions.
1. calculate the heat lost by the hot water in the calorimeter.
initial temp: 22.90 C
finial temp: 20.25 C

2. calculate the heat gained by the cold water.
initial temp: 22.23 C
finial temp: 39.55 C

he didnt give a mass for the water, he said use 1.00 g/mL as its density and 4.18 J/gK as its specific heat.

these are the equations he gave us to use:

qmetal=-(qwater+qcalorimeter)

q=m C dT

i think he wants us to use the first equation, but im not sure how to do that.
Title: Re: Heat lost and Heat gained
Post by: rabolisk on March 24, 2011, 06:59:34 PM: There's more to these questions than you are giving us. At the least we need the volume of water, heat of calorimeter, etc.
Title: Re: Heat lost and Heat gained
Post by: stag on March 24, 2011, 08:01:53 PM: This is for the first question.
50mL of cold water at temp: 22.23 C
50mL of hot water at temp: 63.2 C
hot and cold water temp: 39.55 C
density: 1.00 g/mL
specific heat capacity: 4.18 J/gK
Title: Re: Heat lost and Heat gained
Post by: opti384 on March 24, 2011, 09:25:51 PM: What are you waiting for? You have all the data now and what's left is just applying the figures and numbers into the equation
Title: Re: Heat lost and Heat gained
Post by: stag on March 24, 2011, 10:03:12 PM: (4.18)(50)(23.25-22.90)=73.15
did i use the right temperatures?
Title: Re: Heat lost and Heat gained
Post by: opti384 on March 24, 2011, 10:15:57 PM: Quote
(4.18)(50)(23.25-22.90)=73.15

Is this for your first question? I was just wondering where the 23.25 came from.
Title: Re: Heat lost and Heat gained
Post by: stag on March 24, 2011, 11:39:09 PM: it came from my second post with the added information.
i think i used the wrong temperatures, so i changed it and this is what i got.
(4.18)(50)(39.55-63.22)=-4942.85

and for the second question i got:
(4.18)(50)(39.55-22.23)=3619.88

is this correct?
Title: Re: Heat lost and Heat gained
Post by: opti384 on March 24, 2011, 11:54:44 PM: There seems to be nothing wrong to me. Just keep in mind that the unit is J.
Title: Re: Heat lost and Heat gained
Post by: stag on March 24, 2011, 11:58:33 PM: How would i calculate the heat gained by the calorimeter using those results? he didnt give a heat capacity for the calorimeter.
Title: Re: Heat lost and Heat gained
Post by: stag on March 25, 2011, 12:55:07 AM: would i add the them together?
like this:
-(-4942.85 J+3619.88 J)=1322.97 J
Title: Re: Heat lost and Heat gained
Post by: opti384 on March 26, 2011, 03:40:28 AM: Quote
would i add the them together? like this:-(-4942.85 J+3619.88 J)=1322.97 J

Yes. When hot water and cold water are mixed, think as the heat from the hot water transferred to the cold water. According to you calculations, hot water lost 4942.85J but cold water gained only 3619.88J.
Title: Re: Heat lost and Heat gained
Post by: stag on March 31, 2011, 07:51:28 PM: So would the calorimeter constant be
1322.97 J/17.32 C=76.38

im not too sure what the temperature is supposed to be.
Title: Re: Heat lost and Heat gained
Post by: opti384 on April 01, 2011, 03:04:23 AM: Calorimeter constant is expressed as :delta:H/ :delta:T. I think the temperature change would be 17.32 C, if you added hot water to cold water.