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krek1111 [17]
3 years ago
10

What is the final temperature when 150.0 g of water at 90.0 °c is added to 100.0 g of water at 30.0 OC? Note that C, of water ca

ncels. -qhot
Chemistry
2 answers:
levacccp [35]3 years ago
8 0

Answer : The final temperature is, 337.8K

Explanation :

Q_{absorbed}=Q_{released}

As we know that,  

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

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of water at 90^oC = 150 g

m_2 = mass of water at 30^oC= 100 g

T_{final} = final temperature = ?

T_1 = temperature of lead = 90^oC=273+90=363K

T_2 = temperature of water = 30^oC=273+30=303K

c_1\text{ and }c_2 = same (for water)

Now put all the given values in equation (1), we get

150\times (T_{final}-363)=-[100\times (T_{final}-303)]

T_{final}=337.8K

Therefore, the final temperature is, 337.8K

ipn [44]3 years ago
6 0

<u>Answer:</u> The final temperature of the mixture is 66°C

<u>Explanation:</u>

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

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 hot water = 150.0 g

m_2 = mass of cold water = 100.0 g

T_{final} = final temperature = ?°C

T_1 = initial temperature of hot water = 90.0°C

T_2 = initial temperature of cold water = 30.0°C

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

Putting values in equation 1, we get:

150\times 4.186\times (T_{final}-90)=-[100\times 4.186\times (T_{final}-30)]\\\\T_{final}=66^oC

Hence, the final temperature of the mixture is 66°C

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