How many joules of heat are removed from a 21.0 g sample of water if it is cooled from 34.0°C
1 answer:
Answer:
527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.
In other words, the sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a change in temperature without there being a change in physical state (solid, liquid or gaseous). The equation that allows to calculate this heat exchange is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT=Tfinal-Tinitial is the change in temperature.
In this case:
- c= 4.184
- m=21 g
- ΔT=Tfinal-Tinitial=28 °C - 34 °C=-6 °C
Replacing:
Q= 4.184 * 21 g* (-6 C)
Q= - 527.184 J
To lower the temperature, heat has to be given, for that the final temperature must be lower than the initial temperature; and it receives the name of transferred heat and has a negative value, as in this case.
<u><em> 527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.</em></u>
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