Measuring specific heat capacity

[Janinever]

I completed high school 9 years ago... please bare with me :)

My problem is with how they calculate the actual answer - this is from an example problem in my textbook.

Homework Statement

A calorimeter cup is made from 0.15kg of Alu and contains 0.20 kg of water. Initially the water and cup have a common temp of 18C. A 0.040 kg mass of unknown material is heated to a temp of 97C and then added to the water. The temp of the water, the cup and unknown material is 22C after thermal equilibriam is reestablished. Ignoring small amount of heat gained by thermometer, find the heat capacity of the unknown material.

Homework Equations

Using the equation that they've explained in the book and taking all the data and putting it in the equation (exactly as in my book) see below :

The Attempt at a Solution

[9.00 x 10^2 J/(kg.C](0.15kg)(4.0C) + [4186 J/(kg.C)](0.20kg)(4.0C)
--------------------------------------------------------------------
(0.040kg)(75.0C)

The answer in the book is 1300 J/(kg.C)


With the calculator I get 1656.266667

So I have no clue how they got to 1300?

Please could someone help me understand how they get to the answer? I understand the rest.

Thanks so much!

 

[zizodev]

hi Janinever, your mistkae is here. you had used specific heat with kelvin not in degree celcuis in this cade you should use the temperture with kelvin also or change the specifi heat constant with degre 9.102 J.Kg-1.°K-1 it's in kelvin k=273+T in(degree celcuis)
sorry for my bad english

 

[Janinever]

Thanks for the reply :) don't understand 100%? :)

 

[zizodev]

i mean pay attention in units, look in this table:

http://data.imagup.com/10/1146563447.jpg

you should change the tempertaure to kelvin or recalculate the specific heat from J/(kg.K to J/(kg.C) you know the temperature in kelvin=273+the temperature in degree celcuis.
.

 

[Janinever]

The entire equation from start to finish is in Celsius - so I'm not sure how converting to K is going to help when the books final answer is in C? Theres no mention or indication of k anywhere in this equation? Could you explain in detail what you mean maybe? Or re read my question just incase we are misunderstanding each other somewhere :)

 

[zizodev]

this is the complete solution :)

http://data.imagup.com/10/1146570192.jpg

 

[gneill]

I completed high school 9 years ago... please bare with me :)

My problem is with how they calculate the actual answer - this is from an example problem in my textbook.

Homework Statement

A calorimeter cup is made from 0.15kg of Alu and contains 0.20 kg of water. Initially the water and cup have a common temp of 18C. A 0.040 kg mass of unknown material is heated to a temp of 97C and then added to the water. The temp of the water, the cup and unknown material is 22C after thermal equilibriam is reestablished. Ignoring small amount of heat gained by thermometer, find the heat capacity of the unknown material.

Homework Equations

Using the equation that they've explained in the book and taking all the data and putting it in the equation (exactly as in my book) see below :

The Attempt at a Solution

[9.00 x 10^2 J/(kg.C](0.15kg)(4.0C) + [4186 J/(kg.C)](0.20kg)(4.0C)
--------------------------------------------------------------------
(0.040kg)(75.0C)

The answer in the book is 1300 J/(kg.C)


With the calculator I get 1656.266667

So I have no clue how they got to 1300?

Please could someone help me understand how they get to the answer? I understand the rest.

Thanks so much!

Could be you're having finger problems with the calculator If I run the same calculation as stated above I obtain 1296 J*kg^-1*K^-1, which rounds nicely to 1300 J*kg^-1*K^-1.

 

[Janinever]

Then I'm definitely doing something wrong because I get the same answer over and over :( Will sit with it a while longer and see if I manage to find the problem. Thank you

 

[Janinever]

Got it! :) Thanks Gneil! I haven't used a scientific calculator in years lol so that explains my horrible mistake with the exponent! Thank you SO MUCH!