Answer:
-26.125 kj
Explanation:
Given data:
Mass of water = 250.0 g
Initial temperature = 30.0°C
Final temperature = 5.0°C
Amount of energy lost = ?
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 = 5.0°C - 30.0°C
ΔT = -25°C
Specific heat of water is 4.18 j/g.°C
Now we will put the values in formula.
Q = m.c. ΔT
Q = 250.0 g × 4.18 j/g.°C × -25°C
Q = -26125 j
J to kJ
-26125 j ×1 kj /1000 j
-26.125 kj
