equation qwater = m × c × ΔT. We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. The water has absorbed the heat of the metal. So‚ qwater = qmetal m=mass of water= density x volume = 1 x 26 = 26 grams ΔT = T(mix) – T(water) = 38.9 – 25.3 = 13.6 q(water) =26 x 13.6 x 4.18 q(water) = 1478 Joules 2. Using the formula qmetal = m × c × ΔT‚ calculate the specific heat of the metal. Use the data from your experiment for the
Premium Specific heat capacity Heat Thermodynamics
Identification of the Specific Heat Capacity for a Calorimeter and of the Enthalpy of an Acid-Base Reaction Abstract The purpose of this lab was to first‚ determine the specific heat capacity of a homemade calorimeter‚ and second‚ to calculate the enthalpy of reaction for an acid-base reaction between 6M KOH and 6M HNO3. To determine the specific heat capacity of the calorimeter‚ two differing temperatures of water were measured and volume was measured and mixed within the calorimeter. The
Free Thermodynamics Enthalpy Heat
PSC 41- Specific Heat Capacity and Latent Heat Constants: Specific heat capacity ofwater=4180 J Kg-1 K-1 Specific heat capacity of ice=2050 J Kg-1 K-1 Specific heat capacity of copper= 386 J Kg-1 K-1 Specific heat capacity of aluminum=900 J Kg-1 K-1 Latent heat of vaporization of water =2.26x106 J Kg-1 Latent heat of fusion of water = 3.34x105 J Kg-1 Density of water =1000 Kg m-3 Conversion : 1L ≡ 103 g for water 1) A 12.5 g sample of an unknown metal‚ heated to 99.0
Premium Thermodynamics Water Heat
The specific latent heat of vaporization of a substance is the quantity of energy required to A. raise the temperature of a unit mass of a substance by one degree Celsius. B. convert a unit mass of liquid to vapour at constant temperature and pressure. C. convert a unit mass of solid to vapour at constant temperature and pressure. D. convert a unit mass of liquid to vapour at a temperature of 100°C and a pressure of one atmosphere.
Free Thermodynamics Temperature Energy
Date: 25th November‚ 2011 Aim: To show that when Hot and Cold water are mixed‚ heat lost by hot water is equal to heat gained by cold water. Theory: This Lab is done to demonstrate “the Law of Conservation of Energy”‚ which implies: - “the heat lost by one (1) substance must be equal to the heat gained by another substance within the system”. Apparatus: Two (2) similar Calorimeters: a Thermometer: a Measuring Cylinder: a Bunsen burner: a Tripod: an Asbestos Gauze Cork Mat. Procedure:
Premium Specific heat capacity Heat Thermodynamics
Honors Laboratory: Specific Heat and Conservation of Energy When energy in the form of heat Q is added to a material‚ the temperature of the material rises. Note that temperature‚ in units of degrees Celsius (°C) or Kelvin (K)‚ is a measure of how hot or cold a substance is‚ while heat‚ in units of joules (J) or calories (cal)‚ is a measure of its thermal energy. 1cal = 4.19J. A measure of the efficiency with which a substance can store this heat energy is known as specific heat capacity‚ or simply the
Premium Heat Temperature Thermodynamics
Experiment I: Thermochemistry Background: Thermochemistry is the branch of chemistry that focuses on the studies of chemical reactions or physical changes and heat associated with chemical reactions. According to the first law of thermodynamics‚ energy cannot be created or destroyed but it can be converted from one form into another and/or transferred between different atoms‚ molecules‚ or substances. In general‚ energy can be classified into two categories: kinetic and potential. Kinetic energy
Premium Energy Thermodynamics Enthalpy
water. To determine the heating value of biodiesel‚ a bomb calorimeter will be utilized to measure the heat produced. The first concept that this method utilizes is an overall energy balance for the fuel and water system. (1) Where is the change in energy of a system‚ [J] is the energy of the fuel‚ [J] is the energy of the water‚ [J] is the heat of the fuel in the system‚ [J] is the heat of the water in the‚ [J] is the change in work of the system‚ [J] is the energy from the transfer
Premium Water Oxygen Heat
the specific latent heat of fusion of water? Specific latent heat of fusion is the amount of heat absorbed or released by a substance when changing states. This could be boiling or melting. Latent heat of fusion is the amount of energy needed to melt a substance‚ while latent heat of vaporisation is the amount of energy needed to boil a substance. The specific latent heat of fusion of water is 334 j/g In the graph‚ you can see that as the temperature increases more energy is needed to heat the
Premium Energy Specific heat capacity Thermodynamics
C=-43.0 C 3. Determine the mass of the water by using the water’s density (specific to the initial temperature) and the volume of the water. Remember‚ density = mass/volume. You can look up the density of the water at your specific temperature at http://www.ncsu.edu/chemistry/resource/H2Odensity_vp.htmﾧl. (5 points) Mass=density*volume=0.9987779*75=75g 4. Use the equation: q = m(SH)ΔT to solve for the amount of heat gained by the
Premium Water Heat Thermodynamics