Studdy AI Logo
Math

QuestionFind the final temperature of a mixture of 250 mL of 25°C water and 100 mL of 95°C water, assuming water density is 1.00 g/mL.
Tfinal=∘CT_{\text{final}} = ^\circ \mathrm{C}

Studdy Solution

STEP 1

1. The specific heat capacity of water is constant at 4.18 J/g⋅∘C4.18 \mathrm{~J/g \cdot ^{\circ}C}.
2. The density of water is 1.00 g/mL1.00 \mathrm{~g/mL}, which allows us to convert volume to mass directly.
3. There is no heat loss to the surroundings, and the system is thermally isolated.
4. The final temperature of the mixture will be somewhere between 25.00∘C25.00^{\circ} \mathrm{C} and 95.00∘C95.00^{\circ} \mathrm{C}.
5. The heat gained by the cooler water is equal to the heat lost by the warmer water because of the conservation of energy.

STEP 2

1. Calculate the mass of each water sample.
2. Set up the heat transfer equation.
3. Solve for the final temperature.

STEP 3

Calculate the mass of the 250.0 mL250.0 \mathrm{~mL} water sample using the density.
Mass of cooler water=250.0 mL×1.00 g/mL \text{Mass of cooler water} = 250.0 \mathrm{~mL} \times 1.00 \mathrm{~g/mL}

STEP 4

Calculate the mass of the 100.0 mL100.0 \mathrm{~mL} water sample using the density.
Mass of warmer water=100.0 mL×1.00 g/mL \text{Mass of warmer water} = 100.0 \mathrm{~mL} \times 1.00 \mathrm{~g/mL}

STEP 5

Set up the heat transfer equation based on the conservation of energy, assuming no heat loss to the environment.
Qlost by warm water=Qgained by cool water Q_{\text{lost by warm water}} = Q_{\text{gained by cool water}}

STEP 6

Express the heat lost by the warmer water in terms of its mass, specific heat capacity, and temperature change.
Qlost=mwarm⋅c⋅(Tinitial warm−Tfinal) Q_{\text{lost}} = m_{\text{warm}} \cdot c \cdot (T_{\text{initial warm}} - T_{\text{final}})

STEP 7

Express the heat gained by the cooler water in terms of its mass, specific heat capacity, and temperature change.
Qgained=mcool⋅c⋅(Tfinal−Tinitial cool) Q_{\text{gained}} = m_{\text{cool}} \cdot c \cdot (T_{\text{final}} - T_{\text{initial cool}})

STEP 8

Combine the heat equations, knowing that Qlost=QgainedQ_{\text{lost}} = Q_{\text{gained}}.
mwarm⋅c⋅(Tinitial warm−Tfinal)=mcool⋅c⋅(Tfinal−Tinitial cool) m_{\text{warm}} \cdot c \cdot (T_{\text{initial warm}} - T_{\text{final}}) = m_{\text{cool}} \cdot c \cdot (T_{\text{final}} - T_{\text{initial cool}})

STEP 9

Cancel out the specific heat capacity cc from both sides of the equation, since it is the same for both samples of water.
mwarm⋅(Tinitial warm−Tfinal)=mcool⋅(Tfinal−Tinitial cool) m_{\text{warm}} \cdot (T_{\text{initial warm}} - T_{\text{final}}) = m_{\text{cool}} \cdot (T_{\text{final}} - T_{\text{initial cool}})

STEP 10

Insert the known values into the equation.
100.0⋅(95.00−Tfinal)=250.0⋅(Tfinal−25.00) 100.0 \cdot (95.00 - T_{\text{final}}) = 250.0 \cdot (T_{\text{final}} - 25.00)

STEP 11

Solve for TfinalT_{\text{final}} by distributing and combining like terms.
9500−100.0⋅Tfinal=250.0⋅Tfinal−6250 9500 - 100.0 \cdot T_{\text{final}} = 250.0 \cdot T_{\text{final}} - 6250

STEP 12

Combine the TfinalT_{\text{final}} terms on one side of the equation and the constants on the other side.
9500+6250=250.0⋅Tfinal+100.0⋅Tfinal 9500 + 6250 = 250.0 \cdot T_{\text{final}} + 100.0 \cdot T_{\text{final}}

STEP 13

Add the constants and combine the TfinalT_{\text{final}} coefficients.
15750=350.0⋅Tfinal 15750 = 350.0 \cdot T_{\text{final}}

STEP 14

Divide both sides by 350.0350.0 to solve for TfinalT_{\text{final}}.
Tfinal=15750350.0 T_{\text{final}} = \frac{15750}{350.0}

STEP 15

Calculate the final temperature.
Tfinal=45.00∘C T_{\text{final}} = 45.00^{\circ} \mathrm{C}
The final temperature of the mixture is 45.00∘C45.00^{\circ} \mathrm{C}.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord