Question

6. A 90.0 gram sample of an unknown solid is heated to 80 °C and placed into a calorimeter containing 120 grams of water at 24 °C. If the final temperature of the solid sample and the water is 32 °C, what is the specific heat of solid?

PLEASE SHOW WORK​

Answer 1

1.02 J/g°C.

Step-by-step explanation:

We can use the equation:

q = m * c * ΔT

where q is the heat absorbed or released, m is the mass of the substance (in grams), c is the specific heat, and ΔT is the change in temperature (in Celsius).

First, we can calculate the heat gained by the water:

q_water = m_water * c_water * ΔT_water

where m_water is the mass of the water (in grams), c_water is the specific heat of water (4.184 J/g°C), and ΔT_water is the change in temperature of the water.

m_water = 120 g

c_water = 4.184 J/g°C

ΔT_water = (32°C - 24°C) = 8°C

q_water = (120 g) * (4.184 J/g°C) * (8°C) = 4009 J

This means that the heat lost by the unknown solid is equal to the heat gained by the water:

q_solid = -q_water

q_solid = -4009 J

Next, we can calculate the change in temperature of the solid:

ΔT_solid = (32°C - 80°C) = -48°C

Now, we can solve for the specific heat of the solid:

q_solid = m_solid * c_solid * ΔT_solid

-4009 J = (90.0 g) * c_solid * (-48°C)

c_solid = -4009 J / (90.0 g * -48°C)

c_solid = 1.02 J/g°C

Therefore, the specific heat of the unknown solid is 1.02 J/g°C.