125 g of dry ice (solid CO2) is dropped into a beaker containing 500 g of 66°C water. The dry ice converts directly to gas, leaving the solution. When the dry ice is gone, the final temperature of the water is 29°C. What is the heat of vaporization of solid CO2? (cwater = 1.00 cal/g×°C)

148cal

To find the heat of vaporization of solid CO2, we can use the equation:

q = m * ΔH

where:
q is the heat transferred,
m is the mass of the substance undergoing the phase change (in this case, the solid CO2),
ΔH is the heat of vaporization.

First, let's find the heat gained or lost by the water during the process.

The heat gained by the water can be calculated using the equation:

qwater = mwater * cwater * ΔT

where:
qwater is the heat gained by the water,
mwater is the mass of the water,
cwater is the specific heat capacity of water,
ΔT is the change in temperature of the water.

Given:
mwater = 500 g
cwater = 1.00 cal/g×°C
ΔT = (29°C - 66°C) = -37°C (negative because the water is cooling down)

Plugging in the values:

qwater = (500 g) * (1.00 cal/g×°C) * (-37°C)
qwater = -18500 cal

Next, we need to find the heat transferred during the phase change of the solid CO2:

qCO2 = mCO2 * ΔHvap

Given:
mCO2 = 125 g
ΔT = the temperature change of the solid CO2 during phase change.

Since the solid CO2 completely vaporizes, ΔT is equal to the difference between the initial and final temperatures of the water:

ΔT = (66°C - 29°C) = 37°C

Plugging in the values:

qCO2 = (125 g) * ΔHvap

Now, since the heat gained or lost by the water is equal to the heat transferred during the phase change of the solid CO2, we can set up the equation:

qwater = qCO2

-18500 cal = (125 g) * ΔHvap

Rearranging the equation to solve for ΔHvap:

ΔHvap = (-18500 cal) / (125 g)
ΔHvap ≈ -148 cal/g

Therefore, the heat of vaporization of solid CO2 is approximately -148 cal/g. Note that the negative sign indicates that heat has been lost during the process.

To find the heat of vaporization of solid CO2, we need to use the principle of energy conservation and the equation for heat transfer. The energy transferred to the water will be equal to the energy required to vaporize the dry ice.

The equation for heat transfer is:

Q = m * c * ΔT

Where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature

First, let's calculate the heat transferred from the water:

Q_water = m_water * c_water * ΔT_water

Where:
m_water = 500 g (mass of water)
c_water = 1.00 cal/g×°C (specific heat capacity of water)
ΔT_water = (29°C - 66°C) = -37°C (change in temperature)

Q_water = 500 g * 1.00 cal/g×°C * -37°C
Q_water = -18,500 cal

The negative sign indicates that heat has been lost by the water.

Next, we can find the heat transferred from the dry ice. Since it goes directly from a solid to a gas, it undergoes sublimation. The energy required for sublimation is the heat of vaporization.

Q_sublimation = m_dry_ice * ΔH_sublimation

Where:
m_dry_ice = 125 g (mass of dry ice)
ΔH_sublimation = Heat of vaporization of solid CO2 (to be determined)

We can set up the equation using the principle of energy conservation:

Q_water + Q_sublimation = 0

Substituting the known values, we get:

-18,500 cal + 125 g * ΔH_sublimation = 0

Solving for ΔH_sublimation:

ΔH_sublimation = 18,500 cal / 125 g
ΔH_sublimation = 148 cal/g

Therefore, the heat of vaporization of solid CO2 (dry ice) is 148 cal/g.