Answer:
2621.75 j heat is required to increase the temperature 25.5°C to 46°C.
Explanation:
Given data:
Mass of sample = 142.1 g
Initial temperature = 25.5°C
Final temperature = 46°C
Specific heat capacity of Al = 0.90 J/g.°C
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
ΔT = 46°C - 25.5°C
ΔT = 20.5°C
Q = 142.1 × 0.90 J/g.°C × 20.5°C
Q = 2621.75 j
Thus, 2621.75 j heat is required to increase the temperature 25.5°C to 46°C.
