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.

To calculate the specific heat (c) of the metal, we can use the formula:

c = (mwater * specific heat of water * (Twater2 - Twater1)) / (mmetal * (Tmetal - Twater2))

where:
- mwater is the mass of water in grams
- specific heat of water is 4.184 J/(g•°C), as given
- Twater1 and Twater2 are the initial and final temperatures of the water in degrees Celsius
- mmetal is the mass of the metal in grams
- Tmetal is the temperature of the metal in degrees Celsius

Substituting the given values:

mwater = ? (not provided)
specific heat of water = 4.184 J/(g•°C)
Twater1 = 26.0 °C
Twater2 = 29.0 °C
mmetal = ? (not provided)
Tmetal = 84.0 °C

Without the mass of water (mwater) and the mass of the metal (mmetal), we cannot perform the calculation to determine the specific heat (c) of the metal. Please provide these missing values to proceed with the calculation.