A physical science student adds 500g of 0 degree C ice to 500g of 80 degree C coffee. What is the final temperature of the drink after thermal equilibrium is reached? How much of the ice was melted by the process?

Assume coffee has same specific heat as water.

To find the final temperature of the drink after thermal equilibrium is reached, we can use the principle of conservation of energy.

The first step is to calculate the amount of heat gained or lost by each substance.

For the ice, we use the formula:

Q = m * c * ΔT

where:
Q is the amount of heat gained or lost,
m is the mass of the substance (in this case, ice),
c is the specific heat capacity of the substance (in this case, the specific heat capacity of ice, which is 2.09 J/g°C),
ΔT is the change in temperature.

Since we're starting with ice at 0°C, and assuming it eventually reaches 0°C again after melting, the change in temperature is 0°C.

Q_ice = m_ice * c_ice * ΔT_ice
= 500g * 2.09 J/g°C * 0°C
= 0 J

Therefore, no heat is gained or lost by the ice during this process.

For the coffee, we use the formula:

Q = m * c * ΔT

where:
Q is the amount of heat gained or lost,
m is the mass of the substance (in this case, coffee),
c is the specific heat capacity of the substance (in this case, the specific heat capacity of water, which is 4.18 J/g°C),
ΔT is the change in temperature.

Since we have a change in temperature from 80°C to the final temperature, we can write the equation as:

Q_coffee = m_coffee * c_water * ΔT

We can rearrange the equation to solve for the final temperature:

ΔT = Q_coffee / (m_coffee * c_water)

Substituting the values into the equation:

ΔT = Q_coffee / (500g * 4.18 J/g°C)
= Q_coffee / 2090 J/g

Now, we need to find the amount of heat gained or lost by the coffee. Since heat gained by the coffee equals heat lost by the ice (due to conservation of energy), we have:

Q_coffee = -Q_ice

Substituting the values:

Q_coffee = -0 J

Therefore, the final temperature of the drink after thermal equilibrium is reached is the same as the initial temperature of the coffee, which is 80°C.

Now, let's find out how much of the ice was melted during this process. The heat required to melt the ice is given by the formula:

Q_melt = m_ice * H_melt

where:
Q_melt is the amount of heat required to melt the ice,
m_ice is the mass of the ice,
H_melt is the heat of fusion of ice, which is 334 J/g.

Substituting the values:

Q_melt = 500g * 334 J/g
= 167,000 J

Since heat gained by the coffee is equal to the heat required to melt the ice, we have:

Q_melt = Q_coffee
167,000 J = 0 J

Therefore, none of the ice was melted during this process.

To summarize:
- The final temperature of the drink after thermal equilibrium is reached is 80°C.
- None of the ice was melted during this process.

Ice heat gain = coffee heat loss.

Assume all the ice melts.

500(g)*80(cal/g)= 500 (g)* 1(cal/gC)*(80-T)
where T is the final temperature.

T = 0

Each gram of melting ice gains absorbs 80 cal of heat while melting. That is enough to lower the temperature of 80 g of water by 1 degree.

In this case, all of the ice melts and the water temperature reaches 0 C.

It the computed T had been less than 0, you would have had to assume that all of the ice did not melt, and do the problem differently.