Asked by Ashley

In the lab you submerge 150 g of 50 ∘C nails in 120 g of 20 ∘C water. (The specific heat capacity of iron is 0.12 cal/g⋅∘C.)
Equate the heat gained by the water to the heat lost by the nails and find the final temperature of the water.

Answers

Answered by Damon
water goes up from 20 to T
iron goes down from 50 to T
120 g * 1 cal/gdegC * (T-20)deg C = 150 * 0.12 *(50-T)
120 T - 2400 = 900 - 18 T
138 T = 3300
T = 23.9 deg C
Answered by Bosnian
Qₙₐᵢₗₛ = Qᵥᵥₐₜₕₑᵣ

because heat lost = - 1 ∙ heat gained:

cₙₐᵢₗₛ ∙ mₙₐᵢₗₛ ∙ ∆tₙₐᵢₗₛ = ( - 1 cal / g °C ) ∙ mᵥᵥₐₜₕₑᵣ ∙ ∆tᵥᵥₐₜₕₑᵣ

0.12 cal / g °C ∙ 150 g ∙ ∆tₙₐᵢₗₛ = ( - 1 cal / g °C ) ∙ 150 g ∙ ∆tᵥᵥₐₜₕₑᵣ

But:

Δtₙₐᵢₗₛ = t𝒻ᵢₙₐₗ - 50°C and Δᵥᵥₐₜₕₑᵣ = t𝒻ᵢₙₐₗ - 20°C; therefore

0.12 cal / g °C ∙ 150 g ∙ ( t𝒻ᵢₙₐₗ - 50°C ) = ( - 1 cal / g °C ) ∙ 120 g ∙ ( t𝒻ᵢₙₐₗ - 20°C )

18 cal / °C ( t𝒻ᵢₙₐₗ - 50°C ) = ( - 120 cal / °C ) ∙ ( t𝒻ᵢₙₐₗ - 20°C )

Divide both sides by 1 cal

18 / °C ( t𝒻ᵢₙₐₗ - 50°C ) = ( - 120 / °C ) ∙ ( t𝒻ᵢₙₐₗ - 20°C )

Multiply both sides by 1°C

( 18 / °C ) ∙ 1° C ( t𝒻ᵢₙₐₗ - 50°C ) = [ ( - 120 / °C ) ∙ 1 °C ] ∙ ( t𝒻ᵢₙₐₗ - 20°C )

18 ( t𝒻ᵢₙₐₗ - 50°C ) = ( - 120 ) ∙ ( t𝒻ᵢₙₐₗ - 20°C )

18 t𝒻ᵢₙₐₗ - 900°C = - 120 t𝒻ᵢₙₐₗ + 2400°C

Add 120 t𝒻ᵢₙₐₗ to both sides

18 t𝒻ᵢₙₐₗ + 120 t𝒻ᵢₙₐₗ - 900°C = 2400°C

Add 900°C to both sides

138 t𝒻ᵢₙₐₗ = 3300°C

t𝒻ᵢₙₐₗ = 3300°C / 138

t𝒻ᵢₙₐₗ = 6 ∙ 550°C / 6 ∙ 23

t𝒻ᵢₙₐₗ = 550°C / 23

t𝒻ᵢₙₐₗ = 23.913‬°C


Δᵥᵥₐₜₕₑᵣ = t𝒻ᵢₙₐₗ - 20°C

Δᵥᵥₐₜₕₑᵣ = 550°C / 23 - 20°C

Δᵥᵥₐₜₕₑᵣ = 550°C / 23 - 460 / 23°C

Δᵥᵥₐₜₕₑᵣ = 90°C / 23


tᵥᵥₐₜₕₑᵣ = 20°C + Δᵥᵥᵥₐₜₕₑᵣ

tᵥᵥₐₜₕₑᵣ = 20°C + 90°C / 23

tᵥᵥₐₜₕₑᵣ = 460°C / 23 + 90°C / 23

tᵥᵥₐₜₕₑᵣ = 550°C / 23

tᵥᵥₐₜₕₑᵣ= 23.913‬°C


Check:

Δᵥᵥₐₜₕₑᵣ = t𝒻ᵢₙₐₗ - 20°C

Δᵥᵥₐₜₕₑᵣ = 550°C / 23 - 20°C

Δᵥᵥₐₜₕₑᵣ = 550°C / 23 - 460° / 23 C

Δᵥᵥₐₜₕₑᵣ = 90°C / 23

Δᵥᵥₐₜₕₑᵣ = 3.913°C


∆tₙₐᵢₗₛ = t𝒻ᵢₙₐₗ - 50°C

∆tₙₐᵢₗₛ = 550°C / 23 - 1150°C / 23

∆tₙₐᵢₗₛ = - 600°C / 23

∆tₙₐᵢₗₛ = - 26.087°C


cₙₐᵢₗₛ ∙ mₙₐᵢₗₛ ∙ ∆tₙₐᵢₗₛ = ( - 1 cal / g °C ) ∙ mᵥᵥₐₜₕₑᵣ ∙ ∆tᵥᵥₐₜₕₑᵣ

0.12 cal / g °C ∙ 150 g ∙ ( - 600°C / 23 ) = ( - 1 cal / g °C ) ∙ 120 g ∙ 90°C / 23

- 0.12 cal / g °C ∙ 150 g ∙ 600°C / 23 = ( - 1 cal / g °C ) ∙ 120 g ∙ 90°C / 23

- 18 cal ∙ 600 / 23 = - 120 cal ∙ 90 / 23

- 10800 cal / 23 = - 10800 cal / 23



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