Specific Heat of Metals 1 a

Name Anthony Cervantes Date 09/28/2023 Class PHYSICS_011 Specific Heat (J/g°C) .7864 .4373 1. Determine the change in temperature of the water ( T Δ water ) for Aluminum. T water+ T final water- T Initial water+27.37-25=2.37© which is change of temperature of the water. 2. Calculate the heat ( q ) gained by the water using the following equation: q water = m water ⋅ DT water ⋅ C water , given C water = 4.184 J/(g  C) =99.8x2.33x 4.184 =972.9223J 3. Determine the change in temperature of the Al ( Δ T Al ). TAI= T initial A1 – T final A1=200-27.37=172.63 Celsius. 4. Remembering that the heat gained by the water is equal to the heat lost by the metal (which is why the q is negative), calculate the specific heat of aluminum. q water =− q metal = m Al ⋅ ΔT Al ⋅ C Al which when solved for the heat capacity is: C Al = q metal ( m metal )( ΔT metal ) . Record your result in the Table. =-972.9223/7.2168 x 172.67 = -0.7808 = 0.7808J/(gc) 5. Repeat the calculation of specific heat capacity for Steel and record your result in the Data Table. =-1686.9552/22.5664 x 170.96 =0.4373 J/(gc) 6. If you have a can made of steel and one made of aluminum, describe what will happen to the temperature of the cans when you pull them out of the refrigerator. Include in your discussion connections to the specific heat of the metals. Aluminum has a greater specific heat capacity compared to steel as shown. It would mean thar aluminum would absorb heat faster than steel hence record a higher rate of change of temperature 2ith the respected time for both. Thus, aluminum will get to room temperature first. 7. Many cooking pans are made out of steel and there are also some made of aluminum. Discuss which type of pan you think would be better. Cooking pans are more efficient and effective when they are capable of absorbing heat at a faster rate. Which will then make aluminum cooking pans more viable as they absorb the heat faster compared to pg. 3 - Specific Heat of Metals © Beyond Labz , all rights reserved