Question

If you have 240.0 mL of water at 25.00 °C and add 100.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

Answer 1

Answer: The final temperature of the mixture is 45.6°C

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

• When volume is 240.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{240.0mL}\\\\\text{Mass of water}=(1g/mL\times 240.0mL)=240g$

• When volume is 100.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{100.0mL}\\\\\text{Mass of water}=(1g/mL\times 100.0mL)=100g$

When two water solutions at different temperature are mixed, the amount of heat released by water present at higher temperature will be equal to the amount of heat absorbed by water present at lower temperature..

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$       ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of water solution 1 = 240 g

$m_2$ = mass of water solution 2 = 100 g

$T_{final}$ = final temperature = ?°C

$T_1$ = initial temperature of water solution 1 = 25°C

$T_2$ = initial temperature of water solution 2 = 95°C

c = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

$240\times 4.186\times (T_{final}-25)=-[100\times 4.186\times (T_{final}-95)]\\\\T_{final}=45.6^oC$

Hence, the final temperature of the mixture is 45.6°C

Answer 2

The final temperature of the mixture is about 45.59°C

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Further explanation

Let's recall the Specific Heat Capacity formula as follows:

$\boxed {Q = m c \Delta t }$

where :

Q = heat energy ( J )

m = mass of object ( kg )

c = specific heat capacity ( J/kg°C )

Δt = change in temperature ( °C )

Let us now tackle the problem!

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Given:

mass of first cup of water = m₁ = 240.0 mL × 1.00 g/mL = 240 g

initial temperature of first cup of water = t₁ = 25.00°C

mass of second cup of water = m₂ = 100.0 mL × 1.00 g/mL = 100 g

initial temperature of second cup of water = t₂ = 95.00°C

Asked:

final temperature of mixture = t = ?

Solution:

We will use Conservation of Energy as follows:

$\texttt{Heat Gained } = \texttt{ Heat Released }$

$m_1 c \Delta t_1 = m_2 c \Delta t_2$

$m_1 ( t - t_1 ) = m_2 ( t_2 - t )$

$240 ( t - 25 ) = 100 ( 95 - t )$

$240 t - 6000 = 9500 - 100t$

$240 t + 100 t = 9500 + 6000$

$340 t = 15500$

$t = 15500 \div 340$

$\boxed{t \approx 45.59^oC}$

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Learn more

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• The Speed of Car : brainly.com/question/568302

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Answer details

Grade: High School

Subject: Mathematics

Chapter: Energy