Question

A \(45-\mathrm{g}\) piece of ice at \(0.0{ }^{\circ} \mathrm{C}\) is added to a sample of water at \(8.0{ }^{\circ} \mathrm{C}\). All of the ice melts and the temperature of the water decreases to \(0.0^{\circ} \mathrm{C}\). How many grams of water were in the sample?

Short answer

The sample contained approximately 449.4 grams of water.

Step-by-step solution

- Calculate the energy needed to melt the ice

The energy required to melt the ice can be calculated using the formula: \[ Q = m_{\text{ice}} \times L_f \]where\( m_{\text{ice}} = 45 \text{ g} \) (mass of the ice)\( L_f = 334 \text{ J/g} \) (latent heat of fusion of ice).Substitute the values in to get: \[ Q = 45 \text{ g} \times 334 \text{ J/g} = 15030 \text{ J}. \]

- Calculate the mass of water cooled to melt the ice

The energy to melt the ice comes from the water cooling down. Using the formula: \[ Q = m_{\text{water}} \times c \times (T_i - T_f) \]where \( Q = 15030 \text{ J} \) (energy required to melt the ice), \( c = 4.18 \text{ J/g} \text{°C} \) (specific heat capacity of water),\( T_i = 8.0 \text{°C} \) (initial temperature of the water),\( T_f = 0.0 \text{°C} \) (final temperature of the water).Rearrange to solve for \( m_{\text{water}} \):\[ m_{\text{water}} = \frac{Q}{c \times (T_i - T_f)} = \frac{15030 \text{ J}}{4.18 \text{ J/g°C} \times (8.0 \text{°C} - 0.0 \text{°C})} \]Calculate the value:\[ m_{\text{water}} = \frac{15030 \text{ J}}{4.18 \times 8 } = \frac{15030 \text{ J}}{33.44} \approx 449.4 \text{ g} \]

Key concepts

Latent Heat of Fusion

Let's start by exploring 'latent heat of fusion'. When a substance changes phase, such as ice melting into water, energy is absorbed without changing the temperature of the substance. This energy is known as latent heat. In the case of melting ice, we specifically refer to it as the 'latent heat of fusion'. For ice, this value is 334 J/g. To calculate the energy needed to melt a portion of ice, use the formula:

• \( Q = m_{\text{ice}} \times L_f \)

Here, \(Q\) is the energy in joules, \(m_{\text{ice}}\) is the mass, and \(L_f\) is the latent heat of fusion.

This concept is crucial in understanding how much energy is needed to change the state of the substance without changing its temperature. In the provided problem, 45 grams of ice requires:

\( Q = 45 \text{ g} \times 334 \text{ J/g} = 15030 \text{ J} \)

So, 15030 Joules is the energy required to melt 45 grams of ice at \(0.0^{\circ} \text{C} \).

Specific Heat Capacity

Next, let's talk about 'specific heat capacity'. This is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. The specific heat capacity of water is 4.18 J/g°C, one of the highest among common substances, which is why water heats up and cools down slowly.

When dealing with temperature changes in water, we use the formula:

• \( Q = m_{\text{water}} \times c \times \Delta T \)

Here, \(Q\) is the energy, \(m_{\text{water}}\) is the mass of the water, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.

In the given problem, the water cools from \(8.0^{\circ} \text{C} \) to \(0.0^{\circ} \text{C} \), and the energy transfer is calculated as:

\( Q = 15030 \text{ J} = m_{\text{water}} \times 4.18 \text{ J/g°C} \times (8.0^{\circ} \text{C} - 0.0^{\circ} \text{C})\).

This formula helps us find out how much heat energy the water lost to melt the ice.

Temperature Change Calculations

Finally, let's piece everything together with 'temperature change calculations'. This involves calculating the mass of water that loses certain heat energy when cooling.

• First, you know the energy required to melt the ice is 15030 J.


• Second, the formula for energy change in water:
  \( Q = m_{\text{water}} \times c \times (T_i - T_f) \).


• Rearranging the formula to solve for \(m_{\text{water}}\):
  
  \( m_{\text{water}} = \frac{Q}{c \times (T_i - T_f)} \)

Plugging in the values:
\( m_{\text{water}} = \frac{15030 \text{ J}}{4.18 \times 8} = \frac{15030 \text{ J}}{33.44} \approx 449.4 \text{ g} \)

So, about 449.4 grams of water was initially in the sample. This calculation lets us understand the relationship between energy transfer and temperature change during phase transitions.