Question

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

Answer 1

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

Answer 2

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

Step-by-step 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* c_1* (T_f-T_1)=-m_2* c_2* (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* c_1* (24.5-88.0)^oC=-155g* 4.184J/g^oC* (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`.