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olasank [31]
4 years ago
10

A piece of metal weighing 57.3 g is heated to a temperature of 88.0°C and is then immersed in 155 g of water at a temperature of

21.53°C. After equilibration the temperature is 24.72°C. If CH2O = 4.184 J/g°C, what is Cmetal?
A) .370 J/g°C

B) .164 J/g°C

C) 1.00 J/g°C

D) 2.11 J/g°C

E) .571 J/g°C
Chemistry
2 answers:
Taya2010 [7]4 years ago
8 0

The answer is E. You must use the formula q=mCDeltaT to solve this equation. You must also use the formula that q(reaction)=q(solution) to solve this problem

Ira Lisetskai [31]4 years ago
4 0

Answer:

The specific heat capacity of the metal piece is 0.571J/g^oC.

Explanation:

The heat given by the hot body(metal pace) is equal to the heat taken by the cold body(water).

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of metal = ?

c_2 = specific heat of water = 4.184 J/g^oC

m_1 = mass of metal = 57.3 g

m_2 = mass of water = 155 g

T_f = final temperature of water = 24.72^oC

T_1 = initial temperature of metal = 88^oC

T_2 = initial temperature of water = 21.53^oC

Now put all the given values in the above formula, we get

57.3g\times c_1\times (24.5-88.0)^oC=-155g\times 4.184J/g^oC\times (24.72-21.53)^oC

c_1=0.571 J/g^oC

The specific heat capacity of the metal piece is 0.571J/g^oC.

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