specific heat...would appreciate any help:)?

Question

I have 2 ques:

-Calculate the final temperature when 50mL of water at 60 degrees C are added to 25mL of water at 25 degrees C.

-A piece of metal weighing 5.10 g at a temperature of 48.6 degrees C was placed in a calorimeter into 20.00mL of water at 22.1 degrees C, and the final equilibrium temperature was found to be 26.8 degrees C. What is the specific heat of the metal?

I thought specific heat=q (heat in J)/ (m)(change in temp)...but I can't seem to use that for this problem...

any help is greatly appreciated!!!

Answer 1

Hope this site helps you with your problem :
http://scienceworld.wolfram.com/physics/SpecificHeat.html

Answer 2

First problem: Assume a zero point of 0 C. Then the first sample has 3000 gram-calories of heat, the second has 625 gram-calories, so the total heat present is 3625 gram-calories. Since the total amount of water is 75 ml, a simple division will give the temperature.

Second problem: you can use the same gimmick. Let x be the specific heat of the metal; the total heat at the beginning is 5.1 times 48.6 times x. Work through the rest of the problem in the same way.

Answer 3

problem 1
by common sense you are adding water to water, so forget all those stupid formulae for now... this simply requires you to calculate a weighted mean.
final temp = (50(60) + 25(25))/ (50+25)
= 48.333... C

problem 2
Ok you know that delta Q= mc delta t right?
And also
Heat released from metal = Heat absorbed by water
so...
(mass of metal)(specific heat of metal)(metal's temperature change) = (mass of water) (specific heat of water) (water's temperature change)

so... and we also know that 20 ml of water is approx 20 grams, if you want to be exact, look up the density of water at 22 degrees, but I doubt they're that different
(5.1)(c)(48.6-26.8) = (20)(4.168)(26.8-22.1)
c = (20) (4.168) (4.7) / (5.1) (21.8)
= 3.524 j/gC

Answer 4

first, there is an error in your formula. try using Heat (in J) = (mass) * (change in Temp [∆T] using ◦C) * (Specific Heat in J/g◦C) change the equation using basic algebra skills.

heat= (mass)(∆T)(SH)

SH= heat/(mass)(∆T) <--that is supposed to look like "heat OVER mass times change in temp"

hope i helped a little

Answer 5

if only u could meet me . i can give u detailed information