Answer:
1000 g
Explanation:
Given data:
Initial temperature of water = 45°C
Final temperature of water = 25°C
Heat released = 20 cal (4184×20 = 83680 j)
Mass of water = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Specific heat capacity of water is 4.184 j/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 = 25°C - 45°C
ΔT = -20°C
83680 j = m × 4.184 j/g.°C ×20°C
83680 j = m × 83.68 j/g
m = 83680 j / 83.68 j/g
m = 1000 g
