Question

A piece of a metal having a mass of 150.0000 grams is heated to 99.3 9C. When the hot metal is submerged in 50.000 grams of water originally at 21.3 oC, the final temperature of the water and metal is 48.4 pC. Determine the heat flow (q) of the water in Joules Record the answer to the correct number of significant figures. Show the numerical set-up(s) and include the units in the set-ups. a) Does the temperature of the water increase, decrease, or remain the same? b) Is the heat flow of the water positive, negative, or zero? c) Does the temperature of the metal increase, decrease, or remain the same? d) Is the heat flow of the metal positive, negative, or zero? If the heat gained by the water is the same heat that is lost by the metal, what is the heat flow of the 150.000 gram sample of metal, in Joules? Record your answer in the box to the correct number of significant figures. Determine the value of the specific heat of the 150.000 gram sample of metal, in J/g*oC. Record your answer, to the correct number of significant figures, in the box. Show the numerical set-up(s) and be sure to include the units in your set-ups.

Answer 1

Answer:

Explanation:

a ) The temperature of water increases from 21.3°C to 48.4°C .

b ) Heat flow of water ( q) is positive because there is rise in temperature .

c ) The temperature of metal decreases from 99.3⁰C to 48.4⁰C .

d ) Heat flow of metal is negative because there is fall in temperature .

e ) heat loss of metal = heat gain by water

heat gain by water ( q )  = 50 x 4.2 x ( 48.4 - 21.3 )

= 5691 J

q = 5690 J . ( 3 significant figures )

heat loss of metal = mass x specific heat x fall in temp = 5691

150 x specific heat x ( 99.3 - 48.4 ) = 5691

specific heat = 5691 / (150 x 50.9 )

= .745 J / g C   ( three significant figures )