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.
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
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
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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