how many gras of ice at -14 degree celsius are needed to pull 200g of water from 25 degree celsius to 10 degree celsius

To determine how much ice at -14 degrees Celsius is needed to cool 200 grams of water from 25 degrees Celsius to 10 degrees Celsius, we need to consider the principles of heat transfer and the specific heat capacities involved. This involves calculating the heat lost by the water and the heat gained by the ice as it melts and warms up to 0 degrees Celsius.

Understanding the Heat Transfer

First, let’s break down the problem into manageable parts:

• Calculate the heat lost by the water as it cools from 25°C to 10°C.
• Determine how much heat is required to melt the ice and then warm the resulting water to 0°C.
• Set the heat lost by the water equal to the heat gained by the ice to find the mass of ice needed.

Step 1: Heat Lost by the Water

The formula to calculate the heat lost or gained is:

Q = mcΔT

Where:

• Q = heat (in joules)
• m = mass (in grams)
• c = specific heat capacity (for water, it's approximately 4.18 J/g°C)
• ΔT = change in temperature (final temperature - initial temperature)

For the water cooling from 25°C to 10°C:

• m = 200 g
• c = 4.18 J/g°C
• ΔT = 10°C - 25°C = -15°C

Now, substituting these values into the formula:

Q_water = 200 g * 4.18 J/g°C * (-15°C) = -12540 J

This means the water loses 12,540 joules of energy as it cools down.

Step 2: Heat Gained by the Ice

Next, we need to calculate how much heat is required to melt the ice and then warm the resulting water to 0°C. The process involves two stages:

• Melting the ice at -14°C to water at 0°C.
• Heating the melted ice (now water) from 0°C to 0°C (which is just the melting part, as we will consider the next step later).

For the melting process:

The heat required to melt ice is given by:

Q_melt = m_ice * L_f

Where:

• L_f = latent heat of fusion for ice (approximately 334 J/g)

Now, we also need to account for warming the ice from -14°C to 0°C:

Q_warm = m_ice * c * ΔT

Where:

• ΔT = 0°C - (-14°C) = 14°C
• c = 4.18 J/g°C (specific heat of water, which is the same for ice in terms of warming)

Combining these, the total heat gained by the ice is:

Q_ice = m_ice * L_f + m_ice * c * ΔT

Substituting the values:

Q_ice = m_ice * 334 J/g + m_ice * 4.18 J/g°C * 14°C

Q_ice = m_ice * (334 + 58.52) J/g = m_ice * 392.52 J/g

Step 3: Setting the Heat Lost Equal to Heat Gained

Now we can set the heat lost by the water equal to the heat gained by the ice:

-12540 J = m_ice * 392.52 J/g

Solving for m_ice:

m_ice = -12540 J / 392.52 J/g ≈ 31.98 g

Thus, approximately 32 grams of ice at -14 degrees Celsius are needed to cool 200 grams of water from 25 degrees Celsius to 10 degrees Celsius. This calculation illustrates the principles of energy conservation and the specific heat capacities involved in phase changes and temperature changes.