Can you help with the final row of caluctions on the data tables? thank you

USCTY Data Table 1: Object Description Object Description First Metal Second Metal Mass of water in calorimeter, 25 ml = 25 g [25g ] [25g ] 49.89g/lead 17.67/steel Mass of metal object weight washers Starting temperature of water (room temperature) [24C ] [24C ] Starting temperature of object = (30c ] Highest final temperature of water & object [40C ] Data Table 2: First Metal Objects Mass (E) initial Tfinal AT C (cal/g "C) Water in [25g ] 24d [40c ] [16c ] [ ] calorimeter First metal [49.89g ] 90d 40g [50c ] Data Table 3: Second Metal Objects Mass (E) initial Trimal AT C (cal/g "c) Water in [25g ] 24g 30c [ ] calorimeter Second metal [17.67g ] C 92d 30 62g Displayobject. Calculations 1. Calculate the specific heat of each metal. Q = Q lost by object gained by water Note: Negative Q represents heat lost by metal, and positive Q represents heat gained by water. 2. Solve for the specific heat of each metal (c ). Set up the equation: -myCATm - mucWAT.. m, = mass of the metal C- = specific heat of the metal To = change of temperature of the metal (7, -7) m = mass of water in calorimeter Cw = specific heat of water, 1.00 cal/g 'C AT = change of temperature of the water (T -T) 3. Because you know which metals you have analyzed and have the theoretical specific he values for those metals under Table 1, you should be able to calculate the percent error your experimentally derived value. experimental value - theoretical value % error = * 100 theoretical valueDiscussion and Review When heat energy (Q) is added to a material, the temperature of that material rises. The temperature is measured in degrees Celsius ("C) or in kelvins (K), while the specific heat of that material is measured in calories (cal) or joules (1). The International System of Units (SI) unit for specific heat is joules per kilogram kelvin (1/kg . K). Heat capacity is the proportionality constant between the heat an object absorbs or loses and the resulting temperature change of the object. Specific heat capacity (c) measures the amount of heat needed to increase the temperature of a mass of a material by one degree. The greater a material's specific heat, the more energy must be added to increase its temperature. For example, the specific heat of water is 1.00 cal/g - "C or 4180 J/kg . K. This value means that 1.00 calorie of heat is needed to raise one gram of water by one degree, or 4180 joules of heat is needed to raise one kilogram of water by one degree. According to the law of conservation of energy, when two substances at different temperatures come into contact with one another, heat energy is transferred between them. For example, if you place a piece of hot metal into a container of cold water, the water and its container will become warmer, while the metal will become cooler, until an equilibrium temperature is reached. To measure the specific heat of an unknown substance, a calorimeter is used. In your lab exercise, you will design and use a calorimeter to determine the metals that compose your washers and weights. See Table 1 for a list of specific heat values for various materials. To calculate specific heat, you use the following equation: Q = cmAT Where Q represents heat needed to change temperature; c represents specific heat capacity; m represents mass; and AT represents change in temperature.Q = cmAT Where Q represents heat needed to change temperature; c represents specific heat capacity; m represents mass; and AT represents change in temperature. Table 1. Some specific heats of common materials at room temperature. Specific Heat substance J/kg . K cal/g * *c Water 4180 1.00 Air 1001 0.2391 Styrofoam 1131 0.2701 Copper 386 0.0923 Glass 840 0.20 Lead 128 0.0305 Steel 500 0.107 www.HOLscience.com 5 @Hands-On Labs, Inc