I need help with this experiment. First I need help understanding what this equation is asking: Insturctor wants me to calcultat the heat gaied or lost by water for all three trials qwater = (SHw) (mw)( ATw)  would like to understand this formula and I will share my worksheet where I am stuck.

(Figure 1) Enthalpy diagram illustrating Hesa's law. The net reaction is the same as in Figure 5.21 in the textbook, but here we imagine different reactions in our two-step version. As long as we can write a series of equations that add up to the equation we need, and as long as we know a value for AHF for all intermediate reactions, we can calculate the overall AH Figure thalpy Figure Enthalpy CH₂(g) + 2O₂(g) ΔΗ, = -890 kJ AH₁ = -890 kJ AH₂=-607 kJ CO(g) + 2 H₂O(l) + O₂(g) AH. = -283 kl CH₂(g) + 2O₂(8) AH3= -283 kJ CO₂(g) + 2 H₂O(1) Part A Suppose the overall reaction wore modified to produce 2H₂O(g) rather than 2H₂O(1). Would any of the values of AH in the diagram stay the same? O yes O no Submit Provide Feedback AH₂ = -607 kJ CO(g) + 2 H₂O(1) + O₂(8) Request Answer 4 1 of 1 Next > Pr

thermometer- A 54.2 g sample of polystyrene is put into a calorimeter (see sketch at right) that contains 100.0 g of water. The polystyrene sample starts off at 92.7 °C and the temperature of the water starts off at 21.0 °C. When the temperature of the water stops insulated container changing it's 32.6 °C. The pressure remains constant at 1 atm. water Calculate the specific heat capacity of polystyrene according to this experiment. Be sure your answer is rounded to the correct number of significant digits. sample- a calorimeter J g.°C

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