Evaporation and transpiration
Answer:
c = 1.61 j/g.°C
Explanation:
Given data:
Mass of oil = 9 g
Heat added = 824 j
Initial temperature = 30°C
Final temperature = 87°C
Specific heat of oil = ?
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 = Final temperature - final temperature
ΔT = 87°C - 30°C
ΔT = 57 °C
Q = m.c. ΔT
824 j = 9 g × c × 57 °C
824 j = 513 g. °C × c
c = 1.61 j/g.°C
Hopefully this helps sorry
