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A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59.7 ∘C. Part A What is the mass of the water? Express your answer to two significant figures and include the appropriate units.

Respuesta :

Answer:

The mass of water [tex]m_{w}[/tex] = 39.18 gm

Explanation:

Mass of iron [tex]m_{iron}[/tex] = 32.5 gm

Initial temperature of iron [tex]T_{1}[/tex] = 22.4°c = 295.4 K

Specific heat of iron  [tex]C_{iron}[/tex] = 0.448 [tex]\frac{KJ}{kg K}[/tex]

Mass of water = [tex]m_{w}[/tex]

Specific heat of water  [tex]C_{w} = 4.2 \frac{KJ}{kg K}[/tex]

Initial temperature of water [tex]T_{2}[/tex] = 336 K  

Final temperature after equilibrium [tex]T_{f}[/tex] = 59.7°c = 332.7 K

When iron rod is submerged into water then

Heat lost by water  = Heat gain by iron rod

[tex]m_{w}[/tex] [tex]C_{w}[/tex] ([tex]T_{2}[/tex] - [tex]T_{f}[/tex] ) =  [tex]m_{iron}[/tex] [tex]C_{iron}[/tex] ( [tex]T_{f}[/tex] - [tex]T_{1}[/tex] )

Put all the values in above formula we get

[tex]m_{w}[/tex] × 4.2 × ( 336 - 332.7 ) = 32.5 × 0.448 × ( 332.7 - 295.4 )

[tex]m_{w}[/tex] = 39.18 gm

Therefore the mass of water [tex]m_{w}[/tex] = 39.18 gm

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