Answer:
Q = 27060 j
Explanation:
Given data:
Mass of sample = 120 g
Initial temperature = -10.0 °C
Final temperature = 100 °C
Heat added = ?
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 = T₂ - T₁
ΔT = 100- (-10)
ΔT = 110°C
Q = m.c. ΔT
Q = 120 g × 2.05 g/J.°C × 110°C
Q = 27060 j
