A 36.0−g sample of an unknown metal at 99°C was placed in a constant-pressure calorimeter containing 70.0 g of water at 24.0°C. The final temperature of the system was found to be 28.4°C. Calculate the specific heat of the metal. (The heat capacity of the calorimeter is 12.4 J/°C.)

0.52849 j /g °C

Explanation:

Given the following :

Mass of metal = 36g

Δ Temperature of metal = (28.4 - 99)°C = - 70.6°C

Mass of water = 70g

Δ in temperature of water = (28.4 - 24.0) = 4.4°C

Heat lost by metal = (heat gained by water + heat gained by calorimeter)

Quantity of heat(q) = mcΔT

Where; m = mass of object ; c = specific heat capacity of object

Heat lost by metal:

- (36 × c × - 70.6) = 2541.6c - - - - (1)

Heta gained by water and calorimeter :

(70 × 4.184 × 4.4) + (12.4 × 4.4) = 1288.672 + 54.56 = 1343.232 - - - - (2)

Equating (1) and (2)

2541.6c = 1343.232

c = 1343.232 / 2541.6

c = 0.52849 j /g °C

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.