Acalorimeter contained 75.0 g of water at 16.95 c. a 93.3-g sample of iron at 65.58 c was placed in it, giving a final temperature of 19.68 c for the system. calculate the heat capacity of the calorimeter. specific heats are 4.184 j/g c for h2o and 0.444 j/g c for fe. 

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Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter,

So, the heat capacity of calorimeter is .

metric time was an attempt to create a standard unit of time based off of 10, but it never took off an is not included in the metric system (si)