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A foam cup with negligible specific heat is used as a calorimeter. If you mix 175 g of water at 20.0Ā°C and 125 g of water at 95.0Ā°C, what is the final temperature of the water after it is mixed? Assume no loss of heat to the air or container. (the specific heat of water= 4186 J/Kg.oC)

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hamzaahmeds

Answer:

T = 51.25Ā°C

Explanation:

Applying the law of conservation of energy, we get:

where,

mā‚ = mass of cold water = 175 g

mā‚‚ = mass of hot water = 125 g

T = Final temperature of the mixture = ?

Therefore,

[tex](175\ g)(T-20^oC) = (125\ g)(95^oC-T)\\\\T-20^oC = (0.7143)(95^oC-T)\\\\T(1+0.7143) = 20^oC+67.86^oC\\\\T = \frac{87.86^oC}{1.7143}[/tex]

T = 51.25Ā°C

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