posted by Anonymous on .
Imagine that, in lab, you record the mass of a piece of metal, mmetal, as
Tmetal=84.0 degree C
In another step of this lab, you record the mass of water (mwater) in a coffee cup as
Twater1=26.0 degree C
Twater2=29.0 degree C
With the information above and the specific heat of water, 4.184 J/(g•°C), you calculate the specific heat, c, of the metal.
? J/g times degree C
Your post doesn't make a lot of sense. Record the mass of a metal as 84.0 degrees C? Here's what you would do.
mmetal x specific heat metal x (29.0-84.0) + [mw x 4.184 x (29.0-26.0) = 0 and solve for specific heat metal.
I forgot to add mmetal=45g for the first statement and mmetal=130g for the second one.
No, you didn't have two metals. I might buy 45 g in the first statemenet for mass metal but the 130g must be the mass of the water.
Substitute those numbers where I have mmetal and mw and solve for specific heat metal.
I got .6593 for the specific heat. The numbers are the mass of metal 45g for the 84 degree C and 130g for the 26 and 29 degree C.