Answer:
Q = -3980.9 j
Explanation:
Given data:
Mass of sample = 30 g
Initial temperature = 56.7 °C
Final temperature = 25 °C
Specific heat of water = 4.186 j/g.°C
Amount of heat released = ?
Formula:
Q = m.c.ΔT
Q = heat released
m = mass of sample
c = specific heat of given sample
ΔT = change in temperature
Solution:
ΔT = T2 -T1
ΔT = 25 °C - 56.7 °C = - 31.7°C
Q = m.c.ΔT
Q = 30 g × 4.186 j/g.°C × - 31.7°C
Q = -3980.9 j
