Asked by Greg
If you had 95 g of Aluminum at 98°C and dropped it into a 51 ml of water at 22°C, what'll the final temperature be? The specific heat = 0.22cal/(g°C)
Answers
Answered by
DrBob222
Heat lost by Al = heat gained by water so
heat lost by Al - heat gained by water = 0
I assume the 0.22 cal/g is the secific heat of Al. You will need to look the water specific heat.
[mass Al x specific heat Al x (Tfinal - Tinitial)] + [mass H2O x specific heat H2O x Tfinal - Tinitial)] = 0
Substitute the values and solve for Tf which is the only unknown in the equation. Post your work if you get stuck.
heat lost by Al - heat gained by water = 0
I assume the 0.22 cal/g is the secific heat of Al. You will need to look the water specific heat.
[mass Al x specific heat Al x (Tfinal - Tinitial)] + [mass H2O x specific heat H2O x Tfinal - Tinitial)] = 0
Substitute the values and solve for Tf which is the only unknown in the equation. Post your work if you get stuck.
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