Answer:
105.62 grams of water will change its temperature by 3°C when 1325 Joules of heat is added.
Step-by-step explanation:
Where:
Q is the heat energy (1325 J in this case).
m is the mass of water (what we want to find).
c is the specific heat capacity of water, which is approximately 4.18 J/(g°C).
ΔT is the change in temperature (3°C).
Now, rearrange the formula to solve for mass (m):
m = Q / (c * ΔT)
m = 1325 J / (4.18 J/(g°C) * 3°C)
m ≈ 1325 J / 12.54 J/g
m ≈ 105.62 grams