Answer:
THE SPECIFIC HEAT OF THE METAL IS 0.14 J/g K
Step-by-step explanation:
First, write out the variables in the question
Mass of sample = 18 g
Initial temp of metal = 96°C = 96 + 273 = 369 K
Mass of water = 21 g
Initill temp of water = 24°C = 24 + 273 K = 297 K
final temp of water and metal = 26°C = 26 + 273 K = 299 K
temperature change = (
In calorimetry, there is conservation of energy and heat liberated by one is absorbed by the other. That is heat gained is equals to heat lost.
Heat lost by the metal = Heat gained by water
mc ΔT = mc ΔT
18 * C * ( 299 - 369 ) = 21 * 4.2 * ( 299 - 297)
18 * C * -70 = 21 * 4.2 * 2
-1260 C = 176.4
C = 176.4 / 1260
C = 0.14 J/ g K
The specific heat of the metal is hence 0.14 J/g K