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Suppose object A has double the specific heat and triple the mass of object B. If the same amount of heat is applied to both objects, how will the temperature change of A be related to the temperature change of B? Enter your answer to three significant figures.

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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