1-ELEMENTO BERILIO 𝐵𝑒𝑟𝑖𝑙𝑖𝑜 𝑇𝑓 𝑇𝑖 𝑚 100 𝐶° 23.3𝐶° 20 𝑔 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 × 1534 𝑐𝑎𝑙 𝑐 = 0.434 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 = 0.434 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 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 𝐵𝑒𝑟𝑖𝑙𝑖𝑜 𝑇𝑓 𝑇𝑖 𝑚 50 𝑔 100 𝐶° 27.8𝐶° 1. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑄 = 𝑐 × 50𝑔 × 72.2𝐶° 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 × 2944𝑔. 𝐶° 1280 = 𝑐 × 2944 𝑐 = 0.434 𝑐𝑎𝑙 𝑔×° 𝐶 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 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 =? 𝑐𝑎𝑙 𝑔×° 𝐶 ∆ 𝑇 = 27.8 𝐶° − 20𝐶° = 7.8°𝐶 𝑸 =? 𝑄 = 1 × 200𝑔 × 7.8 = 1560𝑐𝑎𝑙 2- ELEMENTO ORO ORO 20 𝑔 𝑇𝑖 100 𝐶° 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑇𝑓 𝑚 20.2𝐶° 20 𝑔 𝑄 = 𝑐 × 50𝑔 𝑄× =72.2𝐶° 𝑐 × 20𝑔 × 79.8𝐶° 𝑄 = 𝑐 × 3610𝑔. 𝐶° 𝑄 = 𝑐 × 1596𝑔. 𝐶° 𝑸 = 𝒎𝒄𝜟𝒕. 𝒎 = 𝟐𝟎 𝒈 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟎. 𝟐 = 𝟕𝟗. 𝟖𝑪° 𝒄 =? 𝑸 =? 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝑚 = 200 𝑔 ∆ 𝑇 = 20.2 𝐶° − 20𝐶° = 0.2°𝐶 𝑄 = 1 × 200𝑔 × 0.2 = 40 𝑐𝑎𝑙 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 𝑥 1596𝑔. 𝐶° 40 = 𝑐 × 1596 𝑐 = 0.025 𝑐𝑎𝑙 𝑔×° 𝐶 ORO 30 𝑔 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 20.4𝐶° 30 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. = 50𝑔 𝑐 × 30𝑔 × 79.6 𝐶° 𝑄 =𝑄𝑐 × × 72.2𝐶° 𝑄=𝑐× 𝑄 = 𝑐 × 3610𝑔. 𝐶°2338 𝑔. 𝐶° 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 𝑥 2338𝑔. 𝐶° 80 = 𝑐 × 2338 𝒎 = 𝟑𝟎 𝒈 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟎. 𝟒 = 𝟕𝟗. 𝟔 𝑪° 𝑐 = 0.034 𝑐𝑎𝑙 𝑔×° 𝐶 ∆ 𝑇 = 20.4 𝐶° − 20𝐶° = 0.4°𝐶 𝒄 =? 𝑄 = 1 × 200𝑔 × 0.4 = 80 𝑐𝑎𝑙 𝑸 =? ORO 40 𝑔 𝑚 = 200 𝑔 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 20.5𝐶° 40 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 40𝑔××72.2𝐶° 79.5 𝐶° 𝑄𝑄==𝑐𝑐××50𝑔 = 𝑐 × 3180 𝑄=𝑐𝑄 × 3610𝑔. 𝐶° 𝑔. 𝐶° 𝒎 = 𝟒𝟎 𝒈 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟎. 𝟓 = 𝟕𝟗. 𝟓 𝑪° 𝑚 = 200 𝑔 𝒄 =? 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 𝑥 3180𝑔. 𝐶° 100 = 𝑐 × 3180 𝑐 = 0.031 𝑐𝑎𝑙 𝑔×° 𝐶 ∆ 𝑇 = 20.5 𝐶° − 20𝐶° = 0.5°𝐶 𝑸 =? 𝑄 = 1 × 200𝑔 × 0.5 = 100 𝑐𝑎𝑙 ORO 50 𝑔 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 20.6𝐶° 50 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 𝒎 = 𝟓𝟎 𝒈 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟎. 𝟔 = 𝟕𝟗. 𝟒 𝑪° 𝑐 ×× 50𝑔 × 79.4 𝐶° 𝑄 = 𝑐𝑄×=50𝑔 72.2𝐶° 𝑄 = 𝑐 ×𝐶°3970 𝑔. 𝐶° 𝑄 = 𝑐 × 3610𝑔. 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝑚 = 200 𝑔 ∆ 𝑇 = 20.6 𝐶° − 20𝐶° = 0.6°𝐶 𝒄 =? 𝑸 =? 𝑄 = 1 × 200𝑔 × 0.6 = 120 𝑐𝑎𝑙 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 𝑥 3970 𝑔. 𝐶° 120 = 𝑐 × 3970 𝑐𝑎𝑙 𝑐 = 0.030 𝑔×° 𝐶 3-ELEMENTO HIERRO HIERRO 20 𝑔 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 20.9 𝐶° 20 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄= × 20𝑔 × 79.1 𝐶° 𝑄 = 𝑐 × 50𝑔 × 𝑐72.2𝐶° 𝑄 = 𝑐 × 1582 𝑔. 𝐶° 𝑄 = 𝑐 × 3610. 𝐶° 𝑄 = 𝑐 𝑥 1582 𝑔. 𝐶° 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 180 = 𝑐 × 1582 𝒎 = 𝟐𝟎 𝒈 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟎. 𝟗 = 𝟕𝟗. 𝟏 𝑪° ∆ 𝑇 = 20.9 𝐶° − 20𝐶° = 0.9°𝐶 𝒄 =? 𝑸 =? HIERRO 30 𝑔 𝑐𝑎𝑙 𝑐 = 0.114 𝑔×° 𝐶 𝑚 = 200 𝑔 𝑄 = 1 × 200𝑔 × 0.9 = 180 𝑐𝑎𝑙 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 21.3 𝐶° 30 𝑔 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑐 ×× 30𝑔 × 78.7𝐶° 𝑄 = 𝑐𝑄×=50𝑔 72.2𝐶° 𝑄 = 𝑐𝐶°× 2361 𝑔. 𝐶° 𝑄 = 𝑐 × 3610. 𝑸 = 𝒎𝒄𝜟𝒕. 𝑄 = 𝑐 𝑥 2361 𝑔. 𝐶° 260 = 𝑐 × 2361 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝒎 = 𝟑𝟎 𝒈 𝑐𝑎𝑙 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟏. 𝟑 = 𝟕𝟖. 𝟕 𝑪° ∆ 𝑇 = 21.3 𝐶° − 20𝐶° = 1.3°𝐶 𝒄 =? 𝑄 = 1 × 200𝑔 × 1.3 = 260 𝑐𝑎𝑙 𝑸 =? HIERRO 40 𝑔 𝑐 = 0.110 𝑔×° 𝐶 𝑚 = 200 𝑔 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 21.7 𝐶° 40 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑐 ××40𝑔 × 78.3𝐶° 𝑄 = 𝑐𝑄 ×= 50𝑔 72.2𝐶° 𝑄 =𝐶°𝑐 × 3132 𝑔. 𝐶° 𝑄 = 𝑐 × 3610. 𝒎 = 𝟒𝟎 𝒈 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟏. 𝟕 = 𝟕𝟖. 𝟑 𝑪° 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 𝑥 3132 𝑔. 𝐶° 340 = 𝑐 × 3132 𝑐𝑎𝑙 𝒄 =? 𝑚 = 200 𝑔 𝑸 =? ∆ 𝑇 = 21.7 𝐶° − 20𝐶° = 1.7°𝐶 𝑄 = 1 × 200𝑔 × 1.7 = 340 𝑐𝑎𝑙 𝑐 = 0.108 𝑔×° 𝐶 HIERRO 50 𝑔 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 22.1 𝐶° 50 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐𝑄 ×= 50𝑔 72.2𝐶° 𝑐 ××50𝑔 × 77.9𝐶° 𝑄 = 𝑐 × 3610. 𝑄 =𝐶° 𝑐 × 3895 𝑔. 𝐶° 𝑄 = 𝑐 𝑥 3895 𝑔. 𝐶° 𝒎 = 𝟓𝟎 𝒈 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟐. 𝟏 = 𝟕𝟕. 𝟗 𝑪° 420 = 𝑐 × 3895 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝑐𝑎𝑙 𝒄 =? 𝑚 = 200 𝑔 𝑸 =? ∆ 𝑇 = 22.1𝐶° − 20𝐶° = 2.1°𝐶 𝑐 = 0.107 𝑔×° 𝐶 𝑄 = 1 × 200𝑔 × 2.1 = 420 𝑐𝑎𝑙 4- ELEMENTO COBRE COBRE 20 𝑔 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 20.7 𝐶° 20 𝑔 𝑄 = 𝑐 ×𝑄50𝑔 = 𝑐××72.2𝐶° 20𝑔 × 79.3𝐶° 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 × 3610. 𝑄 𝐶° = 𝑐 × 1586 𝑔. 𝐶° 𝑸 = 𝒎𝒄𝜟𝒕. 𝑄 = 𝑐 𝑥 1586 𝑔. 𝐶° 140 = 𝑐 × 1586 𝒎 = 𝟐𝟎 𝒈 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝑐 = 0.088 𝑐𝑎𝑙 𝑔×° 𝐶 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟎. 𝟕 = 𝟕𝟗. 𝟑 𝑪° 𝑚 = 200 𝑔 𝒄 =? ∆ 𝑇 = 20.7𝐶° − 20𝐶° = 0.7°𝐶 𝑸 =? 𝑄 = 1 × 200𝑔 × 0.7 = 140 𝑐𝑎𝑙 COBRE 30 𝑔 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 21.1 𝐶° 30 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑄 =𝑐×𝑄 50𝑔 72.2𝐶° =× 𝑐× 30𝑔 × 78.9𝐶° 𝑄 = 𝑐 × 3610.𝑄𝐶°= 𝑐 × 2367 𝑔. 𝐶° 𝒎 = 𝟐𝟎 𝒈 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟏. 𝟏 = 𝟕𝟖. 𝟗 𝑪° 𝒄 =? 𝑸 =? 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 𝑥 2367 𝑔. 𝐶° 220 = 𝑐 × 2367 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝑚 = 200 𝑔 ∆ 𝑇 = 21.1𝐶° − 20𝐶° = 1.1°𝐶 𝑄 = 1 × 200𝑔 × 1.1 = 220 𝑐𝑎𝑙 𝑐𝑎𝑙 𝑐 = 0.092 𝑔×° 𝐶 COBRE 40 𝑔 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 21.4 𝐶° 40 𝑔 𝑄 =𝑄𝑐 × × 72.2𝐶° = 50𝑔 𝑐 × 40𝑔 × 78.6𝐶° 𝑄 = 𝑐 × 3610. 𝑄 = 𝑐𝐶° × 3144 𝑔. 𝐶° 𝑸 = 𝒎𝒄𝜟𝒕. 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟏. 𝟒 = 𝟕𝟖. 𝟔𝑪° 𝑚 = 200 𝑔 𝒄 =? ∆ 𝑇 = 21.4𝐶° − 20𝐶° = 1.4°𝐶 𝑸 =? 𝑄 = 1 × 200𝑔 × 1.4 = 280 𝑐𝑎𝑙 𝑇𝑖 100 𝐶° 𝑄 = 𝑐 𝑥 3144 𝑔. 𝐶° 280 = 𝑐 × 3144 𝒎 = 𝟒𝟎 𝒈 COBRE 50 𝑔 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑇𝑓 𝑚 21.8𝐶° 50 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑄=𝑐× × 50𝑔 72.2𝐶° 𝑄 50𝑔 =𝑐× × 78.2 𝐶° 𝑄 = 𝑐 × 3610. 𝑄 =𝐶°𝑐 × 3910 𝑔. 𝐶° 𝒎 = 𝟓𝟎 𝒈 𝑐 = 0.089 𝑐𝑎𝑙 𝑔×° 𝐶 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 𝑥 3910 𝑔. 𝐶° 360 = 𝑐 × 3910 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟏. 𝟖 = 𝟕𝟖. 𝟐 𝑪° 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝒄 =? 𝑚 = 200 𝑔 𝑸 =? ∆ 𝑇 = 21.8 𝐶° − 20𝐶° = 1.8°𝐶 𝑐𝑎𝑙 𝑐 = 0.092 𝑔×° 𝐶 𝑄 = 1 × 200𝑔 × 1.8 = 360 𝑐𝑎𝑙 5- ELEMENTO GRAFITO GRAFITO 20 𝑔 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 21.3𝐶° 20 𝑔 𝑄 = 𝑐𝑄×=50𝑔 72.2𝐶° 𝑐 ××20𝑔 × 78.7 𝐶° 𝑄 = 𝑐 × 3610. 𝑄 = 𝐶° 𝑐 × 1574 𝑔. 𝐶° 𝑸 = 𝒎𝒄𝜟𝒕. 𝒎 = 𝟐𝟎 𝒈 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟏. 𝟑 = 𝟕𝟖. 𝟕 𝑪° 𝒄 =? 𝑸 =? 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝑚 = 200 𝑔 ∆ 𝑇 = 21.3𝐶° − 20𝐶° = 1.3°𝐶 𝑄 = 1 × 200𝑔 × 1.3 = 260 𝑐𝑎𝑙 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 𝑥 1574 𝑔. 𝐶° 260 = 𝑐 × 1574 𝑐𝑎𝑙 𝑐 = 0.165 𝑔×° 𝐶 GRAFITO 30 𝑔 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑇𝑖 100 𝐶° 𝑇𝑓 22𝐶° 𝑚 30 𝑔 𝑄 = 𝑐 × 50𝑔 × 72.2𝐶° 𝑄 = 𝑐 × 30𝑔 × 78 𝐶° 𝑄 = 𝑐 × 3610. 𝐶° 𝑄 = 𝑐 × 2340 𝑔. 𝐶° 𝑸 = 𝒎𝒄𝜟𝒕. 𝑄 = 𝑐 𝑥 2340 𝑔. 𝐶° 400 = 𝑐 × 2340 𝒎 = 𝟐𝟎 𝒈 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟐 = 𝟕𝟖 𝑪 𝒄 =? 𝑐𝑎𝑙 𝑐 = 0.170 𝑔×° 𝐶 𝑚 = 200 𝑔 ∆ 𝑇 = 22 𝐶° − 20𝐶° = 2°𝐶 𝑸 =? GRAFITO 40 𝑔 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 1 × 200𝑔 × 2 = 400 𝑐𝑎𝑙 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 22.6𝐶° 40 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 𝑄 = 𝑐 ×𝑄 50𝑔 = 𝑐××72.2𝐶° 40𝑔 × 77.4 𝐶° 𝑄 = 𝑐 × 3610.𝑄𝐶° = 𝑐 × 3096 𝑔. 𝐶° 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝒎 = 𝟒𝟎 𝒈 𝑚 = 200 𝑔 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟐. 𝟔 = 𝟕𝟕. 𝟒 𝑪 ∆ 𝑇 = 22.6 𝐶° − 20𝐶° = 2.6°𝐶 𝒄 =? 𝑄 = 1 × 200𝑔 × 2.6 = 520 𝑐𝑎𝑙 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 𝑥 3096 𝑔. 𝐶° 520 = 𝑐 × 3096 𝑐 = 0.167 𝑐𝑎𝑙 𝑔×° 𝐶 𝑸 =? GRAFITO 50 𝑔 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 23.3𝐶° 50 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 𝒎 = 𝟓𝟎 𝒈 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑄=𝑐× 72.2𝐶° 𝑄 50𝑔 = 𝑐 × 50𝑔 × 76.7 𝐶° 𝑄 = 𝑐 × 3610. 𝑄 =𝐶°𝑐 × 3835 𝑔. 𝐶° 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟑. 𝟑 = 𝟕𝟔. 𝟕 𝑪 𝑚 = 200 𝑔 𝒄 =? ∆ 𝑇 = 23.3 𝐶° − 20𝐶° = 3.3°𝐶 𝑸 =? 𝑄 = 1 × 200𝑔 × 3.3 = 660 𝑐𝑎𝑙 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 𝑥 3835 𝑔. 𝐶° 660 = 𝑐 × 3835 𝑐𝑎𝑙 𝑐 = 0.172 𝑔×° 𝐶 6-ELEMENTO ALUMINIO ALUMINIO 20 𝑔 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑐 ×× 20𝑔 × 78.3 𝐶° 𝑄 = 𝑐𝑄×=50𝑔 72.2𝐶° 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 21.7𝐶° 20 𝑔 𝑄 = 𝑐𝐶°× 1566 𝑔. 𝐶° 𝑄 = 𝑐 × 3610. 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝑸 = 𝒎𝒄𝜟𝒕. 𝑚 = 200 𝑔 𝒎 = 𝟐𝟎 𝒈 ∆ 𝑇 = 21.7 𝐶° − 20𝐶° = 1.7°𝐶 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟏. 𝟕 = 𝟕𝟖. 𝟑 𝑪 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 𝑥 1566 𝑔. 𝐶° 340 = 𝑐 × 1566 𝑐𝑎𝑙 𝑐 = 0.217 𝑔×° 𝐶 𝑄 = 1 × 200𝑔 × 1.7 = 340 𝑐𝑎𝑙 𝒄 =? 𝑸 =? ALUMINIO 30 𝑔 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 22.5𝐶° 30 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 𝑄 =𝑄 𝑐× × 72.2𝐶° =50𝑔 𝑐 × 30𝑔 × 77.5 𝐶° 𝑄 = 𝑐 × 3610. 𝑄 = 𝑐𝐶°× 2325 𝑔. 𝐶° 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝒎 = 𝟑𝟎 𝒈 𝑚 = 200 𝑔 𝒄 =? ∆ 𝑇 = 22.5 𝐶° − 20𝐶° = 2.5°𝐶 𝑸 =? 𝑄 = 1 × 200𝑔 × 2.5 = 500 𝑐𝑎𝑙 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 23.3𝐶° 40 𝑔 𝑄 = 𝑐 × 50𝑔 × 72.2𝐶° 𝑄 = 𝑐 × 40𝑔 × 76.7 𝐶° 𝑄 = 𝑐 × 3610. 𝐶° 𝑄 = 𝑐 × 3068 𝑔. 𝐶° 𝑸 = 𝒎𝒄𝜟𝒕. 𝒎 = 𝟒𝟎 𝒈 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟑. 𝟑 = 𝟕𝟔. 𝟕 𝑪 𝒄 =? 𝑄 = 𝑐 𝑥 2325 𝑔. 𝐶° 500 = 𝑐 × 2325 𝑐𝑎𝑙 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟐. 𝟓 = 𝟕𝟕. 𝟓 𝑪 ALUMINIO 40 𝑔 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑐 = 0.215 𝑔×° 𝐶 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 𝑥 3068 𝑔. 𝐶° 660 = 𝑐 × 3068 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝑚 = 200 𝑔 ∆ 𝑇 = 23.3 𝐶° − 20𝐶° = 3.3°𝐶 𝑸 =? 𝑄 = 1 × 200𝑔 × 3.3 = 660 𝑐𝑎𝑙 𝑐𝑎𝑙 𝑐 = 0.215 𝑔×° 𝐶 ALUMINIO 50 𝑔 𝑇𝑖 100 𝐶° 𝑇𝑓 𝑚 24.1𝐶° 50 𝑔 𝑸 = 𝒎𝒄𝜟𝒕. 1 − 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑄 = 𝑐 × 50𝑔 × 72.2𝐶° 𝑄 = 𝑐 × 50𝑔 × 75.9 𝐶° 𝑄 = 𝑐 × 3610. 𝐶° 𝑄 = 𝑐 × 3795 𝑔. 𝐶° 𝒎 = 𝟓𝟎 𝒈 ∆ 𝑻 = 𝟏𝟎𝟎 − 𝟐𝟒. 𝟏 = 𝟕𝟓. 𝟗 𝑪 3. 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑑𝑎𝑡𝑜𝑠 𝑑𝑒 ∆ 𝑻 𝑄 = 𝑐 𝑥 3795 𝑔. 𝐶° 820 = 𝑐 × 3795 𝑐𝑎𝑙 2. ℎ𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑄 𝑑𝑒𝑙 𝑎𝑔𝑢𝑎 𝒄 =? 𝑚 = 200 𝑔 𝑸 =? ∆ 𝑇 = 24.1 𝐶° − 20𝐶° = 4.1°𝐶 𝑄 = 1 × 200𝑔 × 4.1 = 820 𝑐𝑎𝑙 𝑐 = 0.216 𝑔×° 𝐶 porcentaje de error: CALOR ESPECIFICO Metales Hierro Cobre Grafito Oro Berilio Aluminio masa calor especifico 𝐶𝑒 porcentaje de error media CALOR ESPECIFICO Metales 0.114 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.093 𝑐𝑎𝑙 ⁄𝑔 °𝐶 Hierro Cobre 0,169 𝑐𝑎𝑙 ⁄𝑔 °𝐶 Grafito 0,031 𝑐𝑎𝑙 ⁄𝑔 °𝐶 Oro 0,436 𝑐𝑎𝑙 ⁄𝑔 °𝐶 Berilio 0,215 𝑐𝑎𝑙 ⁄𝑔 °𝐶 Aluminio masa Calor especifico 𝐶𝑒 20𝑔 30𝑔 40𝑔 50𝑔 20𝑔 30𝑔 40𝑔 50𝑔 20𝑔 30𝑔 40𝑔 50𝑔 20𝑔 30𝑔 40𝑔 50𝑔 20𝑔 30𝑔 40𝑔 50𝑔 20𝑔 30𝑔 40𝑔 50𝑔 0.114 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.110 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.108 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.107 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.088 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.092 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.089 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.092 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.165 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.170 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.167 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.172 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.025 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.034 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.031 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.030 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.434 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.434 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.434 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.432 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.217 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.215 𝑐𝑎𝑙 ⁄𝑔 ° 𝐶 0.215 𝑐𝑎𝑙 ⁄𝑔°𝐶 0.216 𝑐𝑎𝑙 ⁄𝑔 °𝐶 porcentaje de error €% 0% 3.5% 5.3% 6.14% 5.37% 0.1% 4.3% 0.1% 2.36% 1.77% 0.2% 1.77% 19.35% 9.7% 0% 3.22% 0.5% 0.5% 0.5% 0.9% 0.9% 0. % 0. % 0.46% Media aritmética (x̄ ) 0.110 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.090 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.169 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.030 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.434 𝑐𝑎𝑙 ⁄𝑔 °𝐶 0.216 𝑐𝑎𝑙 ⁄𝑔 °𝐶 Elemento quimico