The heat lost by the aluminum equals the heat gained by the water.
Look up the specific heat of aluminum, C_a, and write the above statement as an equation. Let C_w be the specific heat of water.
M_a*C_a*(100 - T) = M_w*C_w*(T - 20)
M_a = 41 g
M_w = 71 g
C_w = 1.0 cal/C*g
Solve for the unknown final temperature, T
Let's say I put a very hot 100o C, 41 g cube of aluminum in 71 ml water at room temperature (20o C). What is the temperature of the water after equilibrium occurs
1 answer
1 answer