Question

Suppose you heat a metal object with a mass of 34.9 g to 96.4 oC and transfer it to a calorimeter containing 100.0 g of water at 17.1 oC. The water and metal reach a final temperature of 22.6 oC. What is the specific heat of the metal in J/g-oC

Answer 1

Answer:

The specific heat capacity of the metal = 0.894 J/g.°C

Explanation:

With the right assumption that the calorimeter is well insulated and no heat is lost to the calorimeter,

Heat lost by the metal = Heat gained by water

Heat lost by the metal = mCΔT

m = mass of metal = 34.9 g

C = specific heat capacity of the metal = ?

ΔT = change in temperature of the metal = (final temperature) - (Initial temperature)

= 96.4 - 22.6 = 73.8°C

Heat lost by metal = 34.9 × C × 73.8 = (2,575.62C) J

Heat gained by water = mCΔT

m = mass of water = 100 g

C = Specific heat capacity of water = 4.186 J/g.°C

ΔT = change in temperature = 22.6 - 17.1 = 5.5°C

Heat gained by water = 100 × 4.186 × 5.5 = 2,302.3 J

Heat lost by the metal = Heat gained by water

2,575.62C = 2,302.3

C = (2302.3/2575.62) = 0.8938818615 = 0.894 J/g.°C

Hope this Helps!!!

Answer 2

The specific heat of the metal object placed in the calorimeter is 0.894 J/g $\rm ^\circ C$.

Specific heat can be defined as the amount of heat required to change the temperature of 1 gram of a substance by 1 degree celsius. The expression can be given by:

Q = mc$\Delta$T

where, Q = heat lost by metal

m = mass of the metal = 34.9 g

$\Delta$T = change in temperature = 96.4 $\rm ^\circ C$ - 22.6 $\rm ^\circ C$

$\Delta$T = 73.8 $\rm ^\circ C$

c = specific heat of metal

The heat lost by metal = mc$\Delta$T

The heat lost by metal = 34.9 $\times$ c $\times$  73.8 $\rm ^\circ C$

The heat lost by metal = 2,575.62 C

The metal and water are in colorimeter, thus

The heat lost by metal = The heat gained by the water.

The heat gained by water:

C for water = 4.186 J/g $\rm ^\circ C$

mass of water = 100 g

$\Delta$T = 22.6 - 17.1 $\rm ^\circ C$

$\Delta$T = 5.5 $\rm ^\circ C$

Heat gained by water = 100 $\times$ 4.186 $\times$  5.5 $\rm ^\circ C$

Heat gained by water = 2,302.3 Joules

since, The heat lost by metal = The heat gained by the water.

2,302.3 J = 2,575.62 C

C = 0.894 J/g $\rm ^\circ C$.

Thus the specific heat of the metal object placed in the calorimeter is 0.894 J/g $\rm ^\circ C$.

For more information about specific heat, refer to the link:

https://brainly.com/question/21041726