Question

50 POINTS

a 6.7g piece of rock boiled to 100.0 degrees celsius is placed in 100.0 mL of water with an initial temperature of 23 degrees celsius. the equilibrium temperature when the rock is added is 45 degrees celsius. what is the specific heat of the rock?

Answer 1

To calculate the specific heat of the rock, you can use the formula for heat transfer: Q = mcΔT, where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity and ΔT is the change in temperature.

In this case, we can assume that the heat lost by the rock is equal to the heat gained by the water. Therefore:

Q(rock) = Q(water)

m(rock)c(rock)(T(final) - T(initial, rock)) = m(water)c(water)(T(final) - T(initial, water))

where m(rock) = 6.7 g, T(initial, rock) = 100.0°C, T(final) = 45°C, m(water) = 100.0 g (assuming the density of water is 1 g/mL), c(water) = 4.18 J/g°C (specific heat capacity of water), and T(initial, water) = 23°C.

Substituting these values into the equation above and solving for c(rock), we get:

c(rock) = (m(water)c(water)(T(final) - T(initial, water))) / (m(rock)(T(final) - T(initial, rock)))

c(rock) = (100.0 g * 4.18 J/g°C * (45°C - 23°C)) / (6.7 g * (45°C - 100.0°C))

c(rock) ≈ 1.26 J/g°C

So the specific heat of the rock is approximately 1.26 J/g°C.

Answer 2

To solve this problem, we can use the equation:

q = m * c * ΔT

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

In this case, the heat released by the rock is equal to the heat absorbed by the water, so we can write:

q_rock = -q_water

where q_rock is the heat released by the rock and q_water is the heat absorbed by the water.

The heat released by the rock can be calculated as:

q_rock = m_rock * c_rock * ΔT

where m_rock is the mass of the rock and c_rock is the specific heat of the rock. We know that the mass of the rock is 6.7 g and the ΔT is 45 - 100 = -55 degrees Celsius (because the rock is losing heat to the water).

The heat absorbed by the water can be calculated as:

q_water = m_water * c_water * ΔT

where m_water is the mass of the water and c_water is the specific heat of water. We know that the mass of the water is 100.0 g (which is equivalent to 100.0 mL) and the ΔT is 45 - 23 = 22 degrees Celsius (because the water is gaining heat from the rock).

Since q_rock = -q_water, we can set the two equations equal to each other and solve for c_rock:

m_rock * c_rock * ΔT = -m_water * c_water * ΔT

c_rock = -m_water * c_water * ΔT / (m_rock * ΔT)

Plugging in the values, we get:

c_rock = -(100.0 g) * (4.184 J/g°C) * (22°C) / [(6.7 g) * (-55°C)]

c_rock = 0.811 J/g°C

Therefore, the specific heat of the rock is 0.811 J/g°C.