1) a piece of metal weighing 5.10 g at a temperature of 48.6 Cwas placed in a ca
ID: 687661 • Letter: 1
Question
1) a piece of metal weighing 5.10 g at a temperature of 48.6 Cwas placed in a calorimeter into 20.00 ml of water at 22.1 C, andthe final equilibrium temperature was found to be 28.2 C. what isthe specific heat of the metal? 2) describe how could you determine the specific heat of ametal by using the apparatus and techniques in thisexperiment. 1) a piece of metal weighing 5.10 g at a temperature of 48.6 Cwas placed in a calorimeter into 20.00 ml of water at 22.1 C, andthe final equilibrium temperature was found to be 28.2 C. what isthe specific heat of the metal? 2) describe how could you determine the specific heat of ametal by using the apparatus and techniques in thisexperiment.Explanation / Answer
1)know that heat lost byone = heat gained by other.
so here heat gained/lost is = mc(T2-T1)
where, m is the mass
c is the specific heat capacity i.e 4.186 J/K-gm in case ofwater
T2-T1 is the temperature change occuring
we know that density of water is 1 g/cm3
so mass of water = 20 gm for 20 ml
so heat gained by water at 22.1 degrees C = 20 *4.186*(28.2 -22.1)
= 510.692 J
heat lost by metal at 48.6 degrees C = 5.1*c*(48.6-28.2)
= 104.04*c J
now 104.04 c = 510.692
c = 4.9086 J/K-gm
so specific heat capacity of metal is 4.9086J/K-gm
2)here in thecalorimeter
we know that heat gained by one element/substance is equals to theheat lost by other
as in calorimeter no heat exchange takes place.
and we know that specific heat capacity of a metal is definedas
the amount of heat required to change the temperature of metal of 1gm by 1 degrees celcius.
so knowing the heat we can find the specific heat capacity as wedid here.
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