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oN ) Q = 6.48 > 32.07 Specific Heat Calculationé Worksheet Name: Chemistry — Date: ¥ YOU MUST SHOW YOUR WORK TO RECEIVE CREDIT!!!!*#* Here is a chart of specific heat capacities for your use: Material Specific Heat Material Specific Heat Capacity (J/g-°C) Capacity (J/g-°C) aluminum (Al) 0.9025 lead (Pb) 0.1276 concrete 0.84 mercury (Hg) 0.13950 ethyl alcohol (CH,;,CH,OH) 2.4194 rhodium (Rh) 0.2427 gold (Au) 0.12905 solid steel 0.4494 helium (He) 5.1931 titanium (Ti) 0.5526 hydrogen (H,) 14.304 * vanadium (V) 0.4886 iron (Fe) 0.4494 water (H,0) 4.184 1) In order to make tea, 322,000 J of energy were added to 1000.0 g of water. What was the temperature change of the water? [77.0 °C] Qs StoaAT g= 392,006 J \ 2 32a,000%(4 184 J \:le06,04 (AT 4.184 Fe \ = "0 / | m= 5000.6 422,000 T _ e AT= 1.45? VPS AT= J) 4BLT 1000.06, 3% 2) A 32.07 gram sample of vanadium was heated to 75.00 °C (its initial temperature). It was then dumped into a calorimeter. The initial temperature of the calorimeter’s water was 22.50 °C, After the metal was allowed to release all its heat to the calorimeter’s water, 26.30 °C was the final temperature. What mass of distilled water was inthe calorimeter? [484rg] 47, #g SYSTEM ( - 766.60 ) Q=~ 166-0 s= oO. & YY 2%, m= 3A.07g Thon = 2G +3G°C Tisiiat= TE00°C aT= 48.70°C 49,70 = 797.935 166.0 SURROUNDINGS (+766. oy) Q= + 6.0 s= 4.184. Tqee Q:Smh4T m= a a vy Tom = 0G. B6%C SAT Tisitiat™ 98 5OC 5OPC AaT= 3+86°C
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