Cool Coffee with Melting Ice Cube: Solve for T

In summary, the question asks for the change in temperature (in Celsius) of a 190 cm3 insulated Thermos containing hot coffee at 83.0°C after a 18.0 g ice cube at 0°C has melted and reached equilibrium. The coffee is treated as pure water and there are no energy exchanges with the environment. The specific heat of water is 4186 J/kg·K and the latent heat of fusion is 333 kJ/kg. The density of water is 1.00 g/cm3. The equation used to solve this problem is 190 * 4.186 * (83.0 - T) = 18.0 * 333 + 18.0 * 4.186
  • #1
aub
21
0

Homework Statement


An insulated Thermos contains 190 cm3 of hot coffee at 83.0°C. You put in a 18.0 g ice cube at its melting point to cool the coffee. By how many degrees (in Celsius) has your coffee cooled once the ice has melted and equilibrium is reached? Treat the coffee as though it were pure water and neglect energy exchanges with the environment. The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg. The density of water is 1.00 g/cm3.

(the ice is initially at 0 C)


The Attempt at a Solution


i tried this but didnt work :
190 * 4.186 * (83.0 - T) = 18.0 * 333 + 18.0 * 4.186 * (T -0) and solved for T.

can someone help please?
 
Last edited:
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  • #2
Why 3333?

Pay attention to units.
 
  • #3
Borek said:
Why 3333?

Pay attention to units.

i edited my previous post because i wrote 3333 here by mistake..

btw i tried both 333 and .333 and didnt work
 
  • #4
Other than that equation looks OK to me.
 
  • #5


Based on the given information, we can use the principle of conservation of energy to solve for the final temperature (T) of the coffee once the ice has melted. This can be done by equating the energy gained by the ice (due to the latent heat of fusion) to the energy lost by the coffee (due to its decrease in temperature).

Let's first calculate the energy gained by the ice cube:

Q_ice = m_ice * L_fusion (where Q is the energy gained, m is the mass of the ice, and L_fusion is the latent heat of fusion)

Q_ice = 18.0 g * 333 kJ/kg = 5.994 kJ

Next, let's calculate the energy lost by the coffee:

Q_coffee = m_coffee * c * (T_final - T_initial) (where c is the specific heat of water, and T_final and T_initial are the final and initial temperatures of the coffee, respectively)

We can rearrange this equation to solve for T_final:

T_final = T_initial - (Q_coffee / (m_coffee * c))

Plugging in the values given in the problem, we get:

T_final = 83.0°C - (5.994 kJ / (190 cm3 * 1.00 g/cm3 * 4.186 J/g·K))

T_final = 83.0°C - (0.00794 K) = 82.992°C

Therefore, the coffee has cooled by approximately 0.008°C once the ice has melted and equilibrium is reached.
 

Related to Cool Coffee with Melting Ice Cube: Solve for T

1. What is the purpose of the melting ice cube in "Cool Coffee with Melting Ice Cube: Solve for T"?

The melting ice cube serves as a way to cool down the temperature of the coffee. As the ice cube melts, it absorbs heat from the coffee, causing the overall temperature to decrease.

2. How does the temperature of the coffee change as the ice cube melts?

The temperature of the coffee decreases as the ice cube melts. This is due to the transfer of heat from the coffee to the melting ice cube, which causes the coffee to lose energy and lower its temperature.

3. What other factors can affect the temperature of the coffee besides the melting ice cube?

The temperature of the coffee can also be affected by the initial temperature of the coffee, the temperature of the room, and the materials used for the cup and the ice cube. The rate at which the ice cube melts can also impact the change in temperature.

4. How can we calculate the final temperature of the coffee after the ice cube has completely melted?

The final temperature of the coffee can be calculated using the specific heat capacities of the coffee and the ice cube, their initial temperatures, and the amount of each substance. This can be done using the formula Q = m x c x ΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

5. Can this experiment be replicated with different types of liquids and materials?

Yes, this experiment can be replicated with different types of liquids and materials. However, the results may vary depending on the specific heat capacities and initial temperatures of the substances used. Additionally, the rate of heat transfer may also differ based on the materials used for the cup and the ice cube.

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