H209 First Law of Thermodynamics - 3.1 Experiment 1 - Specific Heat Capacity of Brass (Solution)
Uploaded by hima · 3 June 2023
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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>= ்భା்మ 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
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