20 𝑔 𝐵𝑒𝑟𝑖𝑙𝑖𝑜 𝑇𝑓 𝑇𝑖 𝑚 100 𝐶° 23.3𝐶° 20 𝑔 procedimiento para hallar el calor especifico de berilio 20 g 1. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑸 = 𝒎𝒄𝜟𝒕. 𝒎 = 𝟐𝟎 𝒈 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟑. 𝟑 = 𝟕𝟔. 𝟕 𝑪° 𝑄 = 𝑐 × 20𝑔 × 76.7 𝐶° 𝑄 = 𝑐 × 1534𝑔/𝐶° 𝒄 =? 𝑸 =? procedimiento para hallar Q en función de los datos de agua: 𝑇𝑖 20𝐶° 𝐴𝑔𝑢𝑎 𝑇𝑓 𝑚 23.3𝐶° 200𝑚𝑙 𝒎 = 200 𝑔 ∆ 𝑻 = 23.3 𝐶° − 20𝐶° = 3.33°𝐶 𝒄=1 𝑸 =? reemplazando Q por c para hallar el calor especifico 𝑄 = 𝑐 × 1534𝑔. 𝐶° 666 = 𝑐 × 2253 𝑐 = 0.295 𝑐𝑎𝑙 𝑔×° 𝐶 1. remplazando los datos: 𝑄 = 1 × 200 × 3.33𝐶° 𝑄 = 666 𝑐𝑎𝑙 30 𝑔 𝐵𝑒𝑟𝑖𝑙𝑖𝑜 𝑇𝑓 𝑇𝑖 𝑚 100 𝐶° 24.9𝐶° 30 𝑔 procedimiento para hallar el calor especifico de berilio 30 g 𝑸 = 𝒎𝒄𝜟𝒕. 2. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝒎 = 𝟑𝟎 𝒈 𝑄 = 𝑐 ×𝑚 ×∆𝑇 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟒. 𝟗 = 𝟕𝟓. 𝟏 𝑪° 𝒄 =? 𝑄 = 𝑐 × 30𝑔 × 75.1 𝐶° 𝑸 =? 𝑄 = 𝑐 × 2253𝑔. 𝐶° procedimiento para hallar Q en función de los datos de agua: 𝒎 = 200 𝑔 ∆ 𝑻 = 24.9 𝐶° − 20𝐶° = 4.9°𝐶 𝒄=1 𝑸 =? 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 1 × 200 × 4.9 𝑄 = 980 𝑐𝑎𝑙 reemplazando Q por c para hallar el calor especifico 𝑄 = 𝑐 × 2253𝑔. 𝐶° 980 = 𝑐 × 2253 𝑐 = 0.4349 𝑐𝑎𝑙 𝑔×° 𝐶 𝐵𝑒𝑟𝑖𝑙𝑖𝑜 𝑇𝑓 𝑇𝑖 𝑚 100 𝐶° 26.4𝐶° 40 𝑔 40 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 1. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑄 = 𝑐 × 40𝑔 × 73.6𝐶° 𝑄 = 𝑐 × 2944𝑔. 𝐶° 𝒎 = 𝟒𝟎 𝒈 𝑄 = 𝑐 × 2944𝑔. 𝐶° 1280 = 𝑐 × 2944 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟔. 𝟒 = 𝟕𝟑. 𝟔𝑪° 𝑚 = 200 𝑔 𝒄 =? ∆ 𝑇 = 26.4 𝐶° − 20𝐶° = 6.4°𝐶 𝑸 =? 𝑄 = 1 × 200 × 6.4 = 1280𝑐𝑎𝑙 50 𝑔 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝐵𝑒𝑟𝑖𝑙𝑖𝑜 𝑇𝑓 𝑇𝑖 𝑚 50 𝑔 100 𝐶° 27.8𝐶° 𝑐 = 0.434 1. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑐𝑎𝑙 𝑔×° 𝐶 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 × 50𝑔 × 72.2𝐶° 𝑸 = 𝒎𝒄𝜟𝒕. 𝑄 = 𝑐 × 3610𝑔. 𝐶° 𝑄 = 𝑐 × 3610𝑔. 𝐶° 𝒎 = 𝟓𝟎 𝒈 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 1560 = 𝑐 × 3610 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟕. 𝟖 = 𝟕𝟐. 𝟐𝑪° 𝑚 = 200 𝑔 𝑐 = 0.432 𝒄 =? ∆ 𝑇 = 27.8 𝐶° − 20𝐶° = 7.8°𝐶 𝑸 =? 20 𝑔 𝑐𝑎𝑙 𝑔×° 𝐶 𝑄 = 1 × 200 × 7.8 = 1560𝑐𝑎𝑙 𝐴𝑙𝑢𝑚𝑖𝑛𝑖𝑜 𝑇𝑓 𝑇𝑖 100 𝐶° 21.7 1. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑚 20 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 𝑄 = 𝑐 × 20𝑔 × 78.3𝐶° 𝑄 = 𝑐 × 1566𝑔. 𝐶° 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝒎 = 𝟐𝟎 𝒈 𝑚 = 200 𝑔 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟏. 𝟕 = 𝟕𝟖. 𝟑𝑪° 𝒄 =? 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 ∆ 𝑇 = 21.7 𝐶° − 20𝐶° = 1.7°𝐶 𝑄 = 𝑐 × 3610𝑔. 𝐶° 340 = 𝑐 × 1566 𝑐𝑎𝑙 𝑐 = 0.217 𝑔×° 𝐶 𝑄 = 1 × 200 × 1.7 = 340𝑐𝑎𝑙 𝑸= 30 𝑔 𝐴𝑙𝑢𝑚𝑖𝑛𝑖𝑜 𝑇𝑓 𝑇𝑖 100 𝐶° 22.5 Q=mcΔt. m=30 g ∆ T=100-22.5=77.5C° c=? Q=? 𝑚 30 𝑔 1. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑄 = 𝑐 × 30𝑔 × 77.5𝐶° 𝑄 = 𝑐 × 2,325𝑔. 𝐶° 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 × 2325𝑔. 𝐶° 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 500 = 𝑐 × 2325 𝑚 = 200 𝑔 ∆ 𝑇 = 22.5𝐶° − 20𝐶° = 2.5°𝐶 𝑐 = 0.215 𝑔×° 𝐶 𝑄 = 1 × 200 × 2.5 = 500𝑐𝑎𝑙 𝑐𝑎𝑙 40 𝑔 𝐴𝑙𝑢𝑚𝑖𝑛𝑖𝑜 𝑇𝑓 𝑇𝑖 100 𝐶° 23.3 1. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑚 40 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 𝒎 = 𝟒𝟎 𝒈 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟑. 𝟑 = 𝟕𝟔. 𝟕𝑪° 𝒄 =? 𝑸 =? 50 𝑔 𝐴𝑙𝑢𝑚𝑖𝑛𝑖𝑜 𝑇𝑓 𝑇𝑖 100 𝐶° 24.1 𝑚 50 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 𝒎 = 𝟓𝟎 𝒈 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟒. 𝟏 = 𝟕𝟓. 𝟗𝑪° 𝒄 =? 𝑸 =? 𝑄 = 𝑐 × 40𝑔 × 77.5𝐶° 𝑄 = 𝑐 × 3100𝑔. 𝐶° 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝑚 = 200 𝑔 ∆ 𝑇 = 23.3𝐶° − 20𝐶° = 3.3°𝐶 𝑄 = 1 × 200 × 3.3 = 660𝑐𝑎𝑙 𝑄 = 𝑐 × 50𝑔 × 75.9𝐶° 𝑄 = 𝑐 × 3795𝑔. 𝐶° 𝑚 = 200 𝑔 ∆ 𝑇 = 24.1𝐶° − 20𝐶° = 4.1°𝐶 𝑄 = 1 × 200 × 4.1 = 820𝑐𝑎𝑙 𝑄 = 𝑐 × 3795𝑔. 𝐶° 820 = 𝑐 × 3795 𝑐𝑎𝑙 𝑐 = 0.216 𝑔×° 𝐶 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 × 3100𝑔. 𝐶° 600 = 𝑐 × 3100 𝑐𝑎𝑙 𝑐 = 0.212 𝑔×° 𝐶