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A metal pellet with a mass of 100.0 g, originally at 116°C, is dropped into a cup of water, initially at
23.8°C. The final temperature is 46.3°C. What is the mass of the water in the cup? The specific heat
of water is 4.184 J/g.°C. The specific heat of the pellet is 0.568 J/g.°C.
A. 56g
B. 48g
C. 42g
D. 100g
E. 0.02g

Respuesta :

Answer:

C, 42g

Explanation:

In thermal equilibrium, both bodies (metal pellet and water) both have the same final temperature (46.3°C).

Assuming no heat is lost to surroundings,

the energy lost from metal pellet = energy gained for water

Since E = mc∆T

(energy = mass x specific heat capacity x temperature change)

mc∆T (metal pellet) = mc∆T (water)

100 x 0.568 x (116-46.3) = m 4.184 (46.3 - 23.8)

3958.96 = 94.14m

m = 42g

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