What temperature will the water reach when 10.1 g cao is dropped into a coffee cup containing 157 g h2o at 18.0°c if the following reaction occurs? cao(s) + h2o(l) → ca(oh)2(s) δh°rxn = –64.8 kj/mol assume that the cup is a perfect insulator and that the cup absorbs only a negligible amount of heat. (the specific heat of water = 4.18 j/g·°c)

Final temperature attained by water = 34.6°C

Explanation:

The reaction of CaO and H₂O is an exothermic reaction. The equation of reaction is given below:

CaO + H₂O > Ca(OH)₂

The quantity of heat given off, ΔH°rxn = 64.8KJ/mol = 64800J/mol

Number of moles of CaO = mass/molar mass, where molar mass of Ca0 = 56g/mol, mass of CaO = 10.1g

Number of moles of CaO = 10.1g/56g/mol =0.179moles

Quantity of heat given off by 0.179 moles = 64800 *0.179 = 11599.2J/mol

Using the formula, Quantity of heat, q = mass * specific heat capacity * temperature rise.

mass of mixture = (10.1 + 157)g = 167.1g, Initial temperature = 18.0°C

Final temperature(T₂) - Initial temperature(T₁) = Temperature rise

11599.2J/mol = 167.1g * 4.18J/g·°C * ( T₂ - 18.0°C)

11599.2 = 698.478T₂ - 12572.604

11599.2 + 12572.604 = 698.478T₂

698.478T₂ = 24171.804

T₂ = 34.6°C

Therefore, final temperature attained by water = 34.6°C

there are two significant figures.