q(Al) + q(HOH) = 0
mcT(Al) + mcT(HOH)= 0
[(25)(0.903)(T-57.4)] + [(25)(4.184)(T-25) = 0
Solve for T.
A hot lump of 25.6 g of aluminum at an initial temperature of 57.4 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water given that the specific heat of aluminum is 0.903 J/(g·°C)? Assume no heat is lost to surroundings.
2 answers
104.3
1 answer