Concrete Engineering vs Thermochemistry

Concrete Engineering Calculator is a professional-grade app with all 56 calculators for concrete and civil engineering — fully unlocked from first launch. Smart auto-fill pickers, calculation history with PDF export, and ACI code references on every calculator. Key Features: Supports both Metric (SI) and Imperial (USCS) units Includes detailed formulas for all calculations Smart auto-fill pickers for concrete mix grades, rebar grades, and bar sizes Calculation history with save, review, and PDF export ACI 318-19 code reference badge on every calculator Beam Calculations: Water/Cementitious Materials Ratio Modulus of Elasticity Tensile Strength Compressive Stress (Beam) Stress in Reinforcement Cross-sectional Area (Tensile and Compressive Reinforcement) Stress (Extreme Fiber of Concrete) Shear and Diagonal Tension Nominal Shear Stress Ultimate Shear Capacity Cracking Moment Bending Moment Capacity (Underreinforced and Rectangular Beams) Column Calculations: Total Allowable Axial Load Spiral Reinforcing Ratio Long-Column Load Reduction Factor (Fixed Ends) Maximum Permissible Eccentricity (Spiral and Tied Columns) Bending Load (Spiral and Tied Columns) Slab and Wall Calculations: Slab Thickness Flexural Stiffness Ratio (Beam Section and Slab Width) Axial Capacity of Wall Nominal Shear Stress on Wall Vertical Shear Reinforcement Ratio of Wall Development Lengths and Crack Control: Tension Development Lengths (Various Bar Types) Compression Development Length Development Length (Hooked Bar) Crack Width Control Additional Calculations: Volume of Spiral Steel to Concrete Core Ratio Moment at Balanced Condition Shear Friction Reinforcement Area Coefficient of Sliding Friction Modular Ratio Steel Ratio Effective Moment of Inertia Total End Shear (Counterfort Walls) Ultimate Strength (No Wind, With Wind, With Earthquake Load) Auto-Fill Reference Pickers: Concrete mix picker fills compressive strength and modulus of elasticity automatically Rebar grade picker fills yield strength and modulus of steel automatically Bar size picker fills cross-sectional area and diameter automatically All pickers support both SI and USCS unit modes

Thermochemistry helps in evaluation of enthalpy or heat release/absorption of a system undergoing numerous temperature changes and phase transitions. The calculations take advantage of known values of heat capacity at constant pressure (Cp) and molar or per gram enthalpy of phase transition. Amount of compound can be defined in grams or moles, in a way that Cp and enthalpy units would match. App provides enthalpy values for each step. Negative enthalpy points to exothermic process – heat release, while positive one to endothermic - heat absorption. °C and K are interchangeable. Calorimetry section provides means for evaluation of the heat capacity of calorimeter and for finding equilibrium temperature of mixed system. Forward arrow button sets the final temperature of the mixture. Backward arrow button sets missing temperature or heat capacity of one of the components. Enthalpy values show heat flow for each component. Example of problems solved by application (screenshots): Problem 1: Calculate the amount of energy required to change 100.0 g of ice at -15.0 °C to steam at 125.0 °C. Known values: Heat of melting = 334.16 J g¯1 Heat of vaporization = 2259 J g¯1 specific heat capacity for solid water (ice) = 2.06 J g¯1 K¯1 specific heat capacity for liquid water = 4.184 J g¯1 K¯1 specific heat capacity for gaseous water (steam) = 2.02 J g¯1K¯1 Solution: 1) Heating of 100.0 g of ice from -15.0°C to 0.0°C: (100.0 g) (15.0 K) (2.06 J g¯1 K¯1) = 3090 J 2) Melting of 100.0 g of ice: (100.0 g) (334.16 J g¯1) = 33416 J 3) Heating of 100.0 g of liquid water from zero to 100.0 Celsius: (100.0 g) (100.0 K) (4.184 J g¯1 K¯1) = 41840 J 4) Evaporations of 100.0 g of liquid: (100.0 g) (2259 J g¯1) = 225900 J 5) Heating of 100.0 g of steam from 100.0 to 125.0 Celsius: (100.0 g) (25.0 K) (2.02 J g¯1 K¯1) = 5050 J 6) Summation of the results: 3090 + 33416 + 41840 + 225900 + 5050 = 309.3 kJ Problem 2: Determine the heat capacity of a coffee-cup calorimeter. During calibration 100.0 g of water at 58.5 °C has been added to 100.0 g of water, already in the calorimeter, at 22.8 °C. Calculate the heat capacity of the calorimeter in J/°C, if final temperature of the water is 39.7 °C. (Specific heat of water is 4.184 J/g °C.) Solution: 1) Heat given up by warm water: q = (100.0 g) (18.8 °C) (4.184 J/g °C) = 7865.92 J 2) Heat absorbed by water in the calorimeter: q = (100.0 g) (16.9 °C) (4.184 J/g °C) = 7070.96 J 3) The difference was absorbed by the calorimeter: 7865.92 - 7070.96 = 794.96 J 4) Calorimeter constant: 794.96 J / 16.9 °C = 47.0 J/°C Problem 3: Determine the final temperature when 10.0 g of aluminum at 130.0 °C mixes with 200.0 grams of water at 25.0 °C. Please note the starting temperature of the metal is above the boiling point of water. In reality, the sample may vaporize a tiny amount of water, but we will assume it does not for the purposes of the calculation. Solution: 1) The colder water will warm up and the warmer metal will cool down. The whole mixture will equilibrate up at the same temperature. The energy which "flowed" out of the warmer metal equals the energy which "flowed" into the colder water: Qaluminum = Qwater (10) (130 - x) (0.901) = (200.0 )(x - 25) (4.18) 117.13 - 0.901x = 83.6x - 2090 x = 26.12 °C. Important! Water didn’t cross temperature of phase transition – vaporization; otherwise calculation would be more complex. Calculation of reaction standard Gibbs free energy: For the general reaction aA + bB -> cC + dD ΔG°rxn = cΔGf°(C) + dΔGf°(D) - aΔGf°(A) - bΔGf°(B) Example: Calculate the Gibbs free energy for the following reaction at 25 °C. Cu (s) + H2O (g) -> CuO (s) + H2 (g) ΔG°rxn = ΔGf°(CuO (s)) – ΔGf°(H2O (g)) = (–129.7 kJ/mol) – (–228.6 kJ/mol) = 98.9 kJ/mol ΔGf° = 0; for elements in their standard state by definition. At equilibrium, ΔG = 0! Important points Application uses dot as a decimal separator. Special attention should be paid for units’ consistency.