Answer:
Explanation:
For problems in calorimetry, there are two basic types, one that includes temperature change and another that does not include temperature change. One should be able to determine which physical event is occurring from context of problem. If a temperature change is specified, one should use q = mcΔT; q = heat flow, m = mass of substance being cooled or heated and ΔT = temperature change occurring.
To fully understand these type problems, one should study The Heating Curve for water (following figure).
Starting with 10g ice at -10⁰C calculate the total thermal heat transfer to warm the sample to steam at +110⁰C. (c(ice) = 0.50cal/g⁰C, c(water) = 1.00cal/g⁰C, c(steam) = 0.48cal/g⁰C, ΔHₓ = heat of fusion = 80 cal/gram, ΔHv = heat of vaporization = 540 cal/gram). Note the first 3 constants have temperature in their dimensions but the ΔH values do not have a temperature term.
There are 5 separate calculations to complete with the total being the sum of all heat flow values determined.
Q₁ is a temperature change region (warming of ice) = mcΔT = (10g)(0.50cal/g·°C)[0°C - (-10°C)]= (10)(.50)(10)cals. = 50 calories
Q₂ is a phase change region (melting of Ice sample) = m·ΔHₓ = (10g)(80cal/g) = 800 calories
Q₃ is a temperature change region (warming of water 0⁰C to 100⁰C) = m·ΔHₓ = (10g)(1.00cal/g⁰C)(100⁰C) = 1000 calories
Q₄ = is a phase change region (vaporization of water) = m·ΔHf =(10g)(540cal/g) = 5400 calories
Q₅ = is a temperature change region (warming of steam 100⁰C to 110⁰C) = mcΔT = (10g)(0.48cal/g⁰C)(10⁰C) 48 calories
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Q(total) = Q₁ + Q₂ + Q₃ + Q₄ + Q₅ = (50 + 800 + 1000 + 5400 + 48)calories = 7,298calories
For joules value, multiply calories by 4.184 joules/calorie.
=> Q(total in joules) = 7,298cals x 4.184 joules/cal = 30,535joules.
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Now, most calorimetry problems are simply fragments of the above. The main issue in solving heat flow calorimetry problems is determining what physical event is occurring; i.e., phase change or temperature change. For phase change, Q = m·ΔHt or for temperature change, Q = mcΔT.
The problem specified is a temperature change problem. Therefore you should use Q = mcΔT and apply the given data. That is,
Q = mcΔT = (80g)(1.00cal/g⁰C)(60⁰C - 40⁰C) = 1,600calories x 4.184joules/cal = 66,944joules or 66.944Kilojoules ≅ 67 Kilojoules (2 sig figs) :-)
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