Question

a calorimeter 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. 

Answer 1

Answer:- $\frac{382.69J}{0C}$ .

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, $\Delta T$ 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:

$q=mc\Delta T$

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:

$q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})$

q = 1901.42 J

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

mass of water is 75.0 g.

$\Delta T$ 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:

$q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)$

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, $q=C.\Delta T$

$C=\frac{q}{\Delta T}$

$C=\frac{1044.746J}{2.73^0C}$

$C=\frac{382.69J}{0C}$

So, the heat capacity of calorimeter is $\frac{382.69J}{0C}$ .