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
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.
2 answers
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
1 answer