Answer:
Q = -4903.14 j
Explanation:
Given data:
Mass of water = 23 g
Initial temperature = 78°C
Final temperature = 27°C
Heat lost = ?
Solution:
Specific heat capacity of water is 4.18 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 = 27°C - 78°C
ΔT = -51°C
Q = 23 g × 4.18 J/g.°C × -51°C
Q = -4903.14 j (negative sign shows heat is released)
