Answered

Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

Find the temperature of a mixture obtained by mixing 32 grams of water at a temperature of 22 degrees Celsius with 88 grams of water at a temperature of 75 degrees Celsius.

Sagot :

asuwafohjames

Answer:

60.87 °C

Explanation:

Applying,

Heat lost = heat gain

cm'(t₂-t₃) = cm(t₃-t₁).............. Equation 1

Equation 1 can futher be simplified to

m'(t₂-t₃) = m(t₃-t₁)................Equation 2

Where m' = mass of the hot water, m = mass of the cold water, t₁ = initial temperature of the cold water, t₂ = initial temperature of the hot water, t₃ = temperature of the mixture.

From the question,

Given: m' = 88 g, m = 32 g, t₁ = 22°C, t₂ = 75°C

Substitute these values into equation 2

88(75-t₃) = 32(t₃-22)

6600-88t₃ = 32t₃-704

32t₃+88t₃ = 6600+704

120t₃ = 7304

t₃ = 60.87 °C

We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.