10g of ice at -20C is dropped into a calorimeter containing 10g of water at 10 C,the specific heat of water is twice that of ice.when the equilibrium is reached the calorimeter will contain??

To solve this problem, we need to analyze the heat transfer that occurs when the ice is added to the warmer water. The key here is to understand how energy moves between the ice and the water until they reach thermal equilibrium. Let's break it down step by step.

Understanding the System

We have two substances: ice and water. The ice is at a temperature of -20°C, and the water is at 10°C. The specific heat of water is twice that of ice, which is an important factor in our calculations. Let’s denote the specific heat of ice as \(c\) and that of water as \(2c\).

Heat Transfer Calculations

When the ice is introduced to the water, the following processes occur:

• The ice will first absorb heat to raise its temperature from -20°C to 0°C.
• Next, the ice will melt into water at 0°C.
• Finally, the resulting water from the melted ice will absorb heat to reach the equilibrium temperature.

Calculating Heat Absorbed by Ice

1. **Heating the Ice:** The heat required to raise the temperature of the ice from -20°C to 0°C can be calculated using the formula:

Q1 = m_ice * c_ice * ΔT

Where:

• m_ice = 10 g
• c_ice = c (specific heat of ice)
• ΔT = 0 - (-20) = 20°C

So, we have:

Q1 = 10g * c * 20°C = 200c

2. **Melting the Ice:** The heat required to melt the ice at 0°C is given by:

Q2 = m_ice * L_f

Where \(L_f\) is the latent heat of fusion for ice. Assuming \(L_f\) is approximately 334 J/g, we get:

Q2 = 10g * 334 J/g = 3340 J

3. **Total Heat Absorbed by Ice:** The total heat absorbed by the ice to reach 0°C and then melt is:

Q_total_ice = Q1 + Q2 = 200c + 3340 J

Calculating Heat Lost by Water

Now, let’s calculate the heat lost by the water as it cools down from 10°C to the equilibrium temperature \(T_f\):

Q_water = m_water * c_water * (T_f - 10°C)

Where:

• m_water = 10 g
• c_water = 2c (specific heat of water)
• ΔT = T_f - 10°C

Thus, we have:

Q_water = 10g * 2c * (T_f - 10°C) = 20c(T_f - 10)

Setting Up the Equation for Equilibrium

At thermal equilibrium, the heat gained by the ice will equal the heat lost by the water:

200c + 3340 J = 20c(T_f - 10)

Solving for Equilibrium Temperature

Now, let’s rearrange the equation to find \(T_f\):

200c + 3340 = 20cT_f - 200c

20cT_f = 200c + 3340 + 200c

20cT_f = 400c + 3340

T_f = (400c + 3340) / 20c

Assuming \(c\) cancels out, we simplify:

T_f = 20 + 167/c

Final Considerations

Since the specific heat of ice is less than that of water, the equilibrium temperature will be above 0°C but below 10°C. The exact value will depend on the specific heat of ice used. However, we can conclude that the calorimeter will contain a mixture of water at the equilibrium temperature, which will be a liquid state since all the ice will have melted.

In summary, the calorimeter will contain water at a temperature slightly above 0°C, with all the initial ice melted into water, resulting in a total of 20g of water at equilibrium.