Answer:
5878.6 j
1.405 cal
Explanation:
Given data:
Specific heat of fat = 0.45 cal/g.°C or 1.9 j/g.°C
Density of fat = 30.94 g/cm³
Initial temperature = 25°C
Final temperature = 35°C
Energy needed = ?
Volume of fat = 10 cm³
Solution:
First of all we will calculate the mass of fat from given density and volume.
d = m/v
30.94 g/cm³ = m/ 10 cm³
m = 30.94 g/cm³ × 10 cm³
m = 309.4 g
Now we will calculate the energy needed for fat.
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 = 35°C - 25°C
ΔT = 10°C
Q = m.c. ΔT
Q = 309.4 g . 1.9 j/g.°C . 10°C
q = 5878.6 j
In calories,
5878.6 / 4184 = 1.405 cal
