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.
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
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
To find the final temperature of the water, you can use the principle of heat transfer. The heat gained by the water is equal to the heat lost by the nails.
To calculate the heat gained by the water, you can use the formula:
Q = m * c * ΔT
Where:
Q is the heat gained (or lost),
m is the mass of the water,
c is the specific heat capacity of water (1 calorie/gram degree Celsius),
ΔT is the change in temperature.
Similarly, to calculate the heat lost by the nails, you can use the same formula. The mass of the nails is given as 150 g, and the specific heat capacity of iron is provided as 0.12 calories/gram degree Celsius.
So, the equation for heat gain by water can be written as:
Q_water = m_water * c_water * ΔT_water
And the equation for heat loss by nails can be written as:
Q_nails = m_nails * c_nails * ΔT_nails
Now, since the heat gained by the water is equal to the heat lost by the nails, we can set these two equations equal to each other:
m_water * c_water * ΔT_water = m_nails * c_nails * ΔT_nails
Substituting the given values:
(120 g) * (1 cal/g⋅∘C) * (final temperature - 20 ∘C) = (150 g) * (0.12 cal/g⋅∘C) * (final temperature - 50 ∘C)
Simplifying the equation, we can solve for the final temperature:
(120) * (final temperature - 20) = (150) * (0.12) * (final temperature - 50)
Now you can solve this equation to find the final temperature of the water.