An unknown material, m1 = 0.45 kg, at a temperature of T1 = 91 degrees C is added to a Dewer (an insulated container) which contains m2 = 1.3 kg of water at T2 = 23 degrees C. Water has a specific heat of cw = 4186 J/(kg⋅K). After the system comes to equilibrium the final temperature is T = 31.4 degrees C.

a. Input an expression for the specific heat of the unknown material.
b. What is the specific heat in J (kgK)?

Explanation:

Let the specific heat of material be s

heat lost by material = m₁ s (T 1 - T ) ,  (T 1 - T ) is fall in temp , m₁ is mass of material

= .45 x s x (91 - 31.4 )

= 26.82 s

Heat gained by water

= m₂ cw (T2 - T )

1.3 x 4186 x ( 31.4 - 23 )

heat lost = heat gained

m₂ cw (T2 - T ) = m₁ s (T 1 - T )

1.3 x 4186 x ( 31.4 - 23 ) =  .45 x s x (91 - 31.4 )

45711.12 = 26.82 s

s = 1704.36

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

spare some change and give it to the poor

explanation:

plz : ((