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Suppose you mix m1 = 22 grams of hot water at T1 = 90 'C with m2 = 82 grams of water at T2 = 20 C. What would be the final equilibrium temperature, Te , of the mixture. The specific heat water is c = 4.18 J/g .C. Please round your answer to one decimal place.The table below shows time, temperature, and temperature change data collected using Pivot Lab#10 (Specific Heat of Water Experiments). The heat supplied, Q, is to be calculated from Q = Pt (where P is the power of the light bulb which is 15 watts and t is the time). For example, when t = 200 s, Q = 15 watts x 200 s = 3000 Joules. Plot Temp. Change (vertical axis) versus Heat Supplied (horizontal axis) and find the slope (use MS Excel). Then calculate the specific heat, c, from c = . The slopexmass mass of the water is 925 grams = 0.925 kg. What is the specific heat of water, C, according to the data? Please make sure that your slope is in three significant figures (from Excel) 1. Use this table and graph to collect and analyze data. Follow the instructions below. Time Temperature Temp. Change Heat Supplied 5 Degree C Celsius Degree Joule, J 22.1 2 200 22.7 0.6 3 400 23.4 1.3 4 600 24.2 2.1 5 800 25 2.9 1000 25.7 3.6