A student mixes 100 mL of 0.50 M NaOH with 100 mL of 0.50 MHCl in a Styrofoam cu
ID: 683811 • Letter: A
Question
A student mixes 100 mL of 0.50 M NaOH with 100 mL of 0.50 MHCl in a Styrofoam cup and observes a temperature increase ofT1. When she repeats this experiment using 200 mLof each solution, she observes a temperature change ofT2. If no heat is lost to the surroundings orabsorbed by the Styrofoam cup, what is the relationship betweenT1 and T2? Please EXPLAIN. A. T2= 4T1 B. T2 = 2T1 C. T2 = 0.5 T1 D. T2 = T1 A student mixes 100 mL of 0.50 M NaOH with 100 mL of 0.50 MHCl in a Styrofoam cup and observes a temperature increase ofT1. When she repeats this experiment using 200 mLof each solution, she observes a temperature change ofT2. If no heat is lost to the surroundings orabsorbed by the Styrofoam cup, what is the relationship betweenT1 and T2? Please EXPLAIN. A. T2= 4T1 B. T2 = 2T1 C. T2 = 0.5 T1 D. T2 = T1 C. T2 = 0.5 T1 D. T2 = T1Explanation / Answer
the temperature changes in the first case due to injecting(through the reaction of HCL with NaOH) a certain amount of energyE into the solution (water). There are 200 ml (so 200g) of water, and the temperature change is T (we'retold). when 200ml (twice the quantity) of each reagent is used,obviously twice as much heat is produced in the reaction. Butthere's also twice as much water, so twice the heat spread overtwice the amount produces the same T. mathematically: let H = the heat emitted when 100ml of 0.5M HCLand 0.5M NaOH are mixed (each sample having M x Vol = 0.5 x 0.1 = 0.05 moles.) then 2H = heat emitted when 200ml of 0.5M HCLand 0.5M NaOH are mixed (each sample now has 0.5 x 0.2 = 0.1 moles) in the first case, this H caused a T according toH = m c T (m is the mass of water, c is its heat capacity) in the second case, H still = m c T, butH is twice what it was, and m is twice what is was (c isunchanged), so T must remain the same for the equation to besatisfiedRelated Questions
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