Question

A 25 gram(m) metal ball is heated to 200C(delta T) with 2330 Joules(q) of energy. What is the specific heat of the metal?

Answer 1

Answer:

The specific heat of the metal is 0.466 $\frac{J}{g*C}$

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

The equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

• Q= 2330 J
• c= ?
• m= 25 g
• ΔT= 200 °C

Replacing:

2330 J= c*25 g* 200 °C

Solving:

$c=\frac{2330 J}{25 g* 200 C}$

c=0.466 $\frac{J}{g*C}$

The specific heat of the metal is 0.466 $\frac{J}{g*C}$

Answer 2

Answer:

$Cp=0.466\frac{J}{g^oC}$

Explanation:

Hello,

In this case, we compute the heat as shown below:

$Q=mCp\Delta T$

Thus, since we know the heat, mass and change in temperature, by solving for the heat capacity we obtain:

$Cp=\frac{Q}{m\Delta T} =\frac{2330J}{25g*200^oC} \\\\Cp=0.466\frac{J}{g^oC}$

Best regards.