Answer:
Q = -200 cal
Explanation:
Given data:
Calories lost = ?
Mass of water = 10 g
Initial temperature = 80°C
Final temperature = 60°C
Solution:
Specific heat capacity of water is 1 cal/g.°C
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 = 60°C- 80°C
ΔT = -20°C
Q = m.c. ΔT
Q = 10 g. 1 cal/g.°C. -20°C
Q = -200 cal
