Answer:
c = 0.064 j/g.°C
Explanation:
Given data:
Mass of metal = 100 g
Initial temperature of metal = 250°C
Volume of water = 400 mL
Initial temperature of water = 15°C
Final temperature = 15.9°C
Heat capacity of metal = ?
Solution:
First of all we will calculate the heat lost by metal into water.
ΔT= 15.9°C - 15°C
ΔT= 0.9 °C
q = mcΔT
q = 400 g× 4.184 j/g.°C × 0.9 °C
q = 1506.24 j
Specific heat capacity of meta;
ΔT = 15.9°C - 250°C
ΔT = -234.1 °C
q = mcΔT
-1506.24 j = 100 g× c × -234.1 °C
(negative sign with q value indicate heat is lost by metal)
-1506.24 j = -23410 g.°C × c
c = -1506.24 j /-23410 g.°C
c = 0.064 j/g.°C
