Answer:
ΔT = 0.78 °C
Explanation:
Given data:
Mass of Al = 9.5 g
Specific heat capacity of Al = 0.9 J/g.°C
Temperature change = ?
Heat added = 67 J
Solution:
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
67 J = 9.5 g × 0.9 j/g.°C × ΔT
67 J = 85.5 j/°C × ΔT
ΔT = 67 J / 85.5 j/°C
ΔT = 0.78 °C
