Answer:
Q = 1147.5 cal
Explanation:
Given data:
Energy absorbed = ?
Mass of water = 75 g
Initial temperature = 22.5°C
Final temperature = 37.8°C
Specific heat of water = 1 cal/g.°C
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 = 37.8°C - 22.5°C
ΔT = 15.3 °C
Q = m.c. ΔT
Q = 75 g . 1 cal/g.°C . 15.3 °C
Q = 1147.5 cal
