[Tybalt]

Is this right?

A 15.5g of chromium, heated to 100 ºC is dropped into 55.5g H2O at 16.5 ºC. The final temp. of the metal and H2O is 18.9 ºC. What is the specific heat capacity of the chromium?

(M1C1∆T2) = - (M2C2∆T2)

[15.5g Cr *? * -81.1 K] = [55.5g H2O * 4.184 * 195.05 K]  - [55.5g H2O * 4.184 * 275.55 K] / [15.5g * 195.05 K] =

- 63986.02 / 2976.78 =

-21.50 J / g ºK

[enahs]

No.


First and foremost, there is no such thing as a negative Kelvin. It does not go below zero.

Your equation is correct:
MC∆T = -MC∆T

You just did the math wrong.
∆T = (T_final-T_initial)

Try the math again.