Answer:
Heat absorbed by water = 3985.26 j
Explanation:
Given data:
Mass of water = 75 g
Initial temperature = 20.0°C
Final temperature = 32.7°C
Specific heat of water = 4.184 j/g.°C
Heat absorbed by water = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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 = 32.7°C - 20°C
ΔT = 12.7 °C
Q = 75 g ×4.184 j/g.°C ×12.7 °C
Q = 3985.26 j
Decantation
Centrifugation
Filtration
Evaporation
