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What is the final temperature when 71.8 g of water at 78.8°C is mixed with 33.6 g of water at 29.0°C? (The specific heat of water is 4.184 J/g·°C.)

User Davmos
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1 Answer

6 votes

Answer:

62.92°

Step-by-step explanation:

given,

initial mass of water, m₁ = 71.8 g

initial temperature, T₁ = 78.8°C

another mass of water, m₂ = 33.6 g

another temperature of water, T₂ = 29° C

Final temperature of mix = ?

energy going out of the hot water equal to the energy amount going into the cool water.


q_(lost) =q_(gain)


m_1 c \Delta T = m_1 c \Delta T


71.8 (78.8-x) = 33.6* (x - 29)


105.4 x = 6632.24

x = 62.92°

Hence, the final temperature of the mix is equal to 62.92°

User Sunny Kumar Aditya
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