Question

Watch Interactive Worked Example Video 9.3 in eText, or look through Example 9.3 on page 376 of textbook. What is different/harder about solving this problem than the example on the previous slide? A. We are not given the final temperature. B. Final temperature appears on both sides of the equation, making it difficult to solve for. C. We have to write one temperature change in terms of the other temperature change because both depend on the unknown variable. D Several steps of algebra are required to get Tr alone on one side of the equation. E All of the above

Answer 1

mass of metal = 76.0 g

mass of water = 64.0 g

temperature of water = 25 oC

temperature of metal = 95.7 oC

here ,

heat loss by metal = heat gain by water

(m Cp dT) metal = (m Cp dT) water

76.0 x Cp x (95.7 - 33.0) = 64.0 x 4.18 x (33 - 25)

Cp = 0.4496

specific heat = 0.450 J / g oC

Note : for above proble , question is not given. in example video