chemistry specific heat?
im going over some chemistry problems before school starts back, but i dont understand this specific heat problem…need help.
a sample of metal with a mass 250g is heated to 130 degrees C and dropped into 425g of water at 26 degrees C. The final temperature of the water is 38.4 degress C. What is the specific heat capacity of the metal if 1.2 x 10 ^3 J are lost to the environment?
PLEASE AND THANK YOU!
q (water) = 425 g x 4.18J x ( 38.4 – 26) = 2.20 x 10^4 J
2.20 x 10^4 – 1.2 x10^3 = 2.08 x 10^4
q ( metal ) = – 2.08 x 10^4 J
q ( metal) = 250 g x specific heat x ( 38.4- 130 ) =
– 2.08 x 10^4
specific heat = 0.908 J / g °C
Specific Heat