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A 45.7 block of an unknown metal at 74.2 °C is placed in 72.9 g of water at 15.9 °C. If the
final temperature of the system is 22.9 °C, what is the specific heat of the metal?

Respuesta :

samueladesida43

Answer:

0.934J/g°C

Explanation:

Using Q = mc∆T

However, in this question;

(Q)water = -(Q)metal

(mc∆T)water = -(mc∆T)metal

According to the information provided in the question;

For water;

m = mass = 72.9g

c = specific capacity of water = 4.184 J/g°C

∆T = 22.9 - 15.9 = 7°C

For metal;

m = mass = 45.7g

c = specific capacity of water = ?

∆T = 22.9 - 72.9 = -50°C

(mc∆T)water = -(mc∆T)metal

(72.9 × 4.184 × 7) = -(45.7 × c × -50)

2135.0952 = -(-2285c)

2135.0952 = 2285c

c = 2135.0952/2285

c of metal = 0.934J/g°C

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