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

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 $0.571J/g^oC$.