11 g of aluminum at 200 C and 19 g of copper are dropped into 55 cm^3 of ethyl alcohol at 15 C. The temperature quickly comes to 24 C. What was the initial temperature of the copper? (answer in C) Express your answer using two significant figures.
Example Problem Statement: An aluminum wing on a passenger jet is 29 m long when its temperature is 29°C. At what temperature would the wing be 7 cm (0.07 m) shorter? Step 1: Write down what you know in symbolic form: The wing is made of aluminum, therefore: α = 25 x 10-6 /oC Length: l = 29 m Temperature: T1 = 29 oC Change in length: Δl = -7 cm (-0.07 m) Step 2: Write down what you don't know in symbolic form: Temperature: T2 = ? oC Step 3: Find an equation that contains what you know and what you don't know: Δl = αlΔT = αl(T2 - T1) Step 4: Solve the equation for what you don't know: Δl = αl(T2 - T1); divide both sides by αl Δl/ αl = (αl/ αl )(T2 - T1) = T2 - T1; add T1 to both sides Δl/αl + T1 = T2 - T1 + T1 = T2 Δl/ αl + T1 = T2 Step 5: Plug in (substitute) what you know and you will find out what you don't know: Δl/ αl + T1 = T2 = (-0.07 m)/( 25 x 10-6 /oC x 29 m) + 29 oC = -96.551 oC + 29 oC = -67.551 oC = -67.6 oC rounded off.
Quantity of heat given by aluminium = mc θ = 0.011* 913*(200 - 24) = 1767.6 J. . Quantity of heat given by copper = mc θ = 0.019* 385*(T - 24) = 7.315 T -175.56 J. . Mass of ethyl alcohol = volume * density = 0.000055 m^3 *789 = 0.043395 kg. Quantity of heat gained by ethyl = mc θ = 0.043395 * 2500*(24 - 15) = 976.3875 J heat lost.= Heat gained 1767.6 + 7.315 T -175.56 = 976.39 7.315 T = -615.65 T = -84.17 ?C This shows that copper also gains heat. ======================================...
