Answer:
Specific heat of metal = 0.9 j/g.°C
Explanation:
Given data:
Mass of metal = 120.7 g
Initial temperature of sample = 90.5°C
Final temperature of sample = 25.7 °C
Heat released = 7030 J
Specific heat of metal = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = final temperature - initial temperature
ΔT = 25.7 °C - 90.5 °C
ΔT = - 64.8°C
Now we will put the values in formula.
Q = m.c. ΔT
7020 J = 120.7 g × c × - 64.8°C
7020 J = 7821.36 g.°C × c
c = 7020 J / 7821.36 g.°C
c = 0.9 j/g.°C
