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Sagot :
Taking into account the definition of calorimetry, the temperature change of A will be one sixth of the temperature change of B.
Calorimetry
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
So, the equation that allows to calculate heat exchanges is:
Q = c× m× ΔT
where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
Temperature change of A related to the temperature change of B
In this case, for object A:
QA = cA× mA× (ΔT)A
and for object B:
QB = cB× mB× (ΔT)B
You know:
Object A has double the specific heat of object B. ⇒ cA= 2× cB
Object A has triple the mass of object B. ⇒ mA= 3× mB
If the same amount of heat is applied to both objects ⇒ QA= QB
Then:
cA× mA× (ΔT)A= QB = cB× mB× (ΔT)B
2× cB × 3× mB× (ΔT)A= cB× mB× (ΔT)B
Solving:
6× cB × mB× (ΔT)A= cB× mB× (ΔT)B
(ΔT)A= (cB× mB× (ΔT)B) ÷ (6× cB × mB)
(ΔT)A=[tex]\frac{1}{6}[/tex] (ΔT)B
Finally, the temperature change of A will be one sixth of the temperature change of B.
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