Final answer:
The final temperature of 25.0 grams of water after absorbing 454 J of heat is approximately 26.34°C, calculated using water's specific heat capacity of 4.184 J/g°C.
Step-by-step explanation:
The final temperature of 25.0 grams of water after absorbing 454 J of heat can be calculated using the specific heat capacity of water, which is 4.184 J/g°C.
The formula to find the temperature change (ΔT) when a certain amount of heat (q) is added is q = mcΔT,
where m is the mass of the water and
c is the specific heat capacity.
So, to find the final temperature (T_f), you would rearrange the formula to ΔT = q/(mc) and then add the ΔT to the initial temperature (T_i), which gives you
T_f = T_i + ΔT.
Using the values provided:
ΔT = 454 J / (25.0 g × 4.184 J/g°C)
= 454 J / 104.6 J/°C
= 4.34°C.
Therefore, the final temperature is 22°C + 4.34°C ≈ 26.34°C.