H209 First Law of Thermodynamics - 3.1 Experiment 1 - Specific Heat Capacity of Brass (Solution)

9749 H2 Physics Experiment (Solution) Topic 8 & 9 : Thermal Physics Page 1 of 4 Name : _____________________ ( ) Class : _____ Date : _________ Experiment No.: 1 Duration: 1 hr Worksheet Title: Specific Heat Capacity of Brass Aim : In this experiment, you are to determine a value for the specific heat capacity of brass. Apparatus : To be shared: 1 x 500 ml beaker with wire gauze at the base of beaker 1 x bunsen burner, tripod, lighter and wire gauze 1 x thermometer (range -10 to 100 degrees celsius) 1 x glass rod 4 x 100 g brass weight Individual: 1 x 250 ml beaker 1 x 250 ml expanded polystrene cup 1 x pair of tongs 1 x ceramic tiles 1 x 100 ml measuring cylinder 1 x thermometer 1 x glass rod (a) (i) Heat some water in the large 500 cm3 beaker and maintain the water at its boiling point. (ii) Immerse the 100 g brass weights into the boiling water. (iii) Measure and record the constant temperature of the brass weights in the boiling water, Tb. Tb = ………………………….. (iv) Assuming that water has a density of 1.00 g cm-3, measure, using the measuring cylinder, 40 g of water at room temperature and pour the water into the polystyrene cup (note that 1 cm -3 = 1 ml). Record the mass mw as well as the temperature Tr of this water. mw = ………….……………… . Tr = …………………..……… [1] (v) As quickly as possible, transfer the one hot brass weight from the boiling water to the water in the polystyrene cup. (vi) Measure and record the highest temperature, Tf reached by the water in the cup. ܶ௙ଵ = 42.0 ℃ ܶ௙ଶ = 42.0 ℃ Average <Tf>= ்೑భା்೑మ ଶ = ସଶ.଴ାସଶ.଴ ଶ = 42.0℃ Tf = …………………..……… [1] 100.0°C 40.0 g 29.0 °C 42.0 °C

9749 H2 Physics Experiment (Solution) Topic 8 & 9 : Thermal Physics Page 2 of 4 (b) Change the mass of water mw at room temperature and repeat steps (a)(ii) to (a)(vi) until you have 6 sets of values for mw and Tf. mw / g Tr / oC Tf1 / oC Tf2 / oC <Tf >/ oC ૚ ࢓࢝ g-1 ࢀࢌ−ࢀ࢘ ࢀ࢈−ࢀࢌ 40.0 29.0 42.0 42.0 42.0 0.0250 0.224 50.0 29.0 39.8 39.7 39.8 0.0200 0.179 60.0 29.0 38.2 38.1 38.2 0.0167 0.149 70.0 29.0 37.1 37.2 37.2 0.0143 0.131 80.0 29.0 36.1 36.1 36.1 0.0125 0.111 90.0 29.0 35.5 35.4 35.5 0.0111 0.101 [7] (c) By assuming that all heat lost by the brass weight is absorbed by the water in the cup, it is suggested that the following equation will relate Tb, Tr and Tf : mwcw (Tf – Tr) = mbcb(Tb – Tf) where mw and cw are respectively the mass and specific heat ca pacity of the water in the cup. mb and cb are the mass and specific heat capacity of the brass weight. Plot a suitable graph to determine if your data supports the above equations. gradient = ……………………. y-intercept = ……………………. [2] Given mwcw (Tf – Tr) = mbcb(Tb – Tf) ൫ࢀࢌ−ࢀ࢘൯ ൫ࢀ࢈−ࢀࢌ൯ =࢓࢈ࢉ࢈ ࢉ࢝ ൬ ૚ ࢓࢝ ൰ By plotting a graph of ൫ࢀࢌିࢀ࢘൯ ൫ࢀ࢈ିࢀࢌ൯ against ቀ ૚ ࢓࢝ ቁ, a straight line graph should be obtained with gradient ࢓࢈ࢉ࢈ ࢉ࢝ and y-intercept 0. For Best Fit Line Using points on best fit line, (0.0260, 0.232) and (0.0120, 0.108), gradient of best fit line = ଴.ଶଷଶି଴.ଵ଴଼ ଴.଴ଶ଺଴ି଴.଴ଵଶ଴ = 8.86 g (3 s.f. follow raw data) y-intercept of best fit line = 0.232 − (8.86)(0.0260) = 0.00164 (3 s.f. follow raw data) For Extreme Fit Line Using points on extreme fit line, (0.0254, 0.220) and (0.0130, 0.120), gradient of extreme fit line = ଴.ଶଶ଴ି଴.ଵଶ଴ ଴.଴ଶହସି଴.଴ଵଷ଴ = 8.06 g (3 s.f. follow raw data) y-intercept of extreme fit line = 0.220 − (8.06)(0.0254) = 0.0153 (3 s.f. follow raw data) Hence, uncertainty in y- intercept = ±|0.00164 − 0.0153| = ±0.01 (1 s.f.) Thus y-intercept = (0.00 ± 0.01) Theoretically, if the above equation is true, the y -intercept of the best fit line should fall near 0. Since the y-intercept of the best fit line falls in the range of 0.00 ± 0.01, the above relationship is most probably valid.

9749 H2 Physics Experiment (Solution) Topic 8 & 9 : Thermal Physics Page 3 of 4 [3] Graph of ൫ࢀࢌିࢀ࢘൯ (ࢀ࢈ିࢀࢌ) against ૚ ࢓࢝ /ࢍି૚ ൫ࢀࢌିࢀ࢘൯ (ࢀ࢈ିࢀࢌ) ૚ ࢓࢝ /ࢍି૚ 0.100 0.120 0.140 0.160 0.180 0.200 0.240 0.220 0.010 0.012 0.014 0.022 0.020 0.016 0.018 0.028 0.024 0.026 (0.0120, 0.108) (0.0260, 0.232) (0.0130, 0.120) (0.0254, 0.220) Line of best fit Line of extreme fit

9749 H2 Physics Experiment (Solution) Topic 8 & 9 : Thermal Physics Page 4 of 4 (d) Comment on any anomalous data or results that you may have been obtained. ………………………………………………………………………………………………….... ……………………………………………… ………………………………………………….... ……………………………………………………………………………………………….. [1] (e) Determine the specific heat capacity cb of the brass weight. Include its appropriate unit. (Take the specific heat capacity of water cw to be 4.20 J g-1 K-1 ) cb = …………………………….. [2] (f) Estimate the uncertainty for the value of cb that you have obtained in (e). Δcb = …… ……… ……………….[1] (g) State one significant source of error that contributed to the uncertainty in this experiment. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ………………………………………………………………………………………………... [1] (h) Suggest improvements that could be made to the experiment to address the errors identified in (g). You may suggest the use of other apparatus or a different procedure. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ………………………………………………………………………………………………... [1] 1) In the process of transferring the brass mass from the beaker to the polystyrene cup, water from the beaker may be transferred over as well, causing systematic errors in Tf recorded. 2) The brass mass provided may not necessarily be of the mass required, causing incorrect values of cb to be calculated. 1) The brass mass could be heated in an oven. 2) Measure and record the value of mb using an electronic pan balance. There is no anomalous point as all the data points lie close to and are evenly scattered about the line of best fit. From the best fit line of ൫்೑ି்ೝ൯ ൫்್ି்೑൯ against ଵ ௠ೢ , ௠್௖್ ௖ೢ would be given by its gradient. ݐ݊݁݅݀ܽݎ݃ ݂݋ ݐݏܾ݁ ݐ݂݅ ݈݁݊݅= ݉௕ܿ௕ ܿ௪ = 8.86 ܿ௕ = (8.86)(4.20) 100 = 0.372 J gିଵ Kିଵ From the extreme line graph of ൫்೑ି்ೝ൯ ൫்್ି்೑൯ against ଵ ௠ೢ , ௠್௖್ ௖ೢ would be given by its gradient. ݐ݊݁݅݀ܽݎ݃ ݂݋ ݁݉݁ݎݐݔ݁ ݐ݂݅ ݈݁݊݅= ݉௕ܿ௕ ܿ௪ = 8.1 ܿ௕ ᇱ = (8.06)(4.20) 100 = 0.339 J gିଵ Kିଵ Hence, Δܿ௕ = ±|ܿ௕ −ܿ௕ ᇱ | = ±|0.372 − 0.339| = ±0.03