Answer:
660J/kg.°C
Explanation:
Using the fact that
∆E of metal + ∆E of water = 0
∆E of the metal = Q = m*c*(T2-T1)
Where
Q = heat lost or gained = ?
m = mass of metal in kg = 0.076kg
c = specific heat of the metal = ?
T1 = 96°C
T2=31°C
∆E of the water=Q= m*c*(T2-T1)
Where
Q = 0.120*4180*6.5 = 3260J
Therefore
3,260J + Q of the metal =0
3,260J + 0.076*C*(31.0–96.0)=0
3,260J - 4.94c = 0
3,260 = 4.94c
3,260/4.94 = c
660J/kg.°C = specific heat of the metal
hope this helps:)
