Question

You placed 43.1 g of an unknown metal at 100 °C into a coffee cup calorimeter that contained 50.0 g of water that was initially at 22.0 °C. The equilibrium temperature of mixing (T0) was determined to be 23.2 °C. The calorimeter constant was known to be 51.5 J/°C. Specific HeatH2O = 4.184 J/g·°Ca. What is the total amount of heat (J) lost by the metal? NG 1.5b. What was the specific heat (J/g·°C) of the metal? NG 1.5

Answer 1

To find the specific heat capacity of the unknown metal, use the formula q = m·c·ΔT, where q is the heat lost/gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Substituting the given values into the formula, we can calculate the specific heat capacity of the metal to be approximately 0.999 J/g·°C.

Step-by-step explanation:

To find the specific heat capacity (K') of the unknown metal, we can use the formula:

q = m·c·ΔT

where q is the heat lost/gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. We can rearrange this formula as:

c = q/(m·ΔT)

Substituting the given values into the formula, we have:

c = (4.18 J/g·°C · (60.15 - 50.0) °C) / (40.0 g · (130.0 - 50.0) °C)

Simplifying the equation and calculating, we find that the specific heat capacity of the unknown metal is approximately K' = 0.999 J/g·°C

Answer 2

(a) The heat released by the metal is -312.48 J

(b) The specific heat of the metal is
`0.0944J/g^oC`

Explanation :

For part A :

Heat released by the metal = Heat absorbed by the calorimeter + Heat absorbed by the water

`q=[q_1+q_2]`

`q=[c_1* \Delta T+m_2* c_2* \Delta T]`

where,

q = heat released by the metal

`q_1` = heat absorbed by the calorimeter

`q_2` = heat absorbed by the water

`c_1` = specific heat of calorimeter =
`51.5J/^oC`

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

`m_2` = mass of water = 50.0 g

`\Delta T` = change in temperature =
`(T_(final)-T_(initial))=23.2-22.0=1.2^oC`

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

`q=[(51.5J/^oC* 1.2^oC)+(50.0g* 4.184J/g^oC* 1.2^oC)]`

`q=312.48J`

Thus, the heat released by the metal is -312.48 J

For part B :

`q=m* c* \Delta T`

q = heat released by the metal = -312.48 J

m = mass of metal = 43.1 g

c = specific heat of metal = ?

`\Delta T` = change in temperature =
`(T_(final)-T_(initial))=23.2-100=76.8^oC`

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

`-312.48J=43.1g* c* 76.8^oC`

`c=0.0944J/g^oC`

Thus, the specific heat of the metal is
`0.0944J/g^oC`