2grams of a substance was dissolved in 75.0 g of water resulting in a temperature decrease from 23.1 ⁰c to 18.7 ⁰c. the specific heat of the solution is 4.18 j/g·⁰c. calculate the amount of heat associated with the water.

1379.4 Joules

Therefore;

Quantity of heat = Mass × specific heat capacity × Change in temperature

Q = mcΔT

In this case;

The substance dissolved in water gained heat while water lost heat energy.

Thus, Heat gained by the substance = heat lost by water

Heat associated with the water  

Mass of water = 75 g

Change in temperature = 4.4°C

Specific heat capacity =  4.18 J/g·⁰C

Heat = mcΔT

        = 75 g × 4.18 J/g·⁰C × 4.4 °C

        =1379.4 Joules

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm