Problem 54 - Chemistry An Atoms First Approach

Question

A biology experiment requires the preparation of a water bath at \(37.0^{\circ} \mathrm{C}\) (body temperature). The temperature of the cold tap water is \(22.0^{\circ} \mathrm{C},\) and the temperature of the hot tap water is \(55.0^{\circ} \mathrm{C} .\) If a student starts with \(90.0 \mathrm{g}\) cold water, what mass of hot water must be added to reach \(37.0^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The student needs to add \(75.0 grams\) of hot water to the \(90.0 grams\) of cold water to achieve a final temperature of \(37.0^{\circ} C\).

Step 1: Write down the specific heat formula and conservation of energy principle

The heat q gained or lost by a substance can be calculated using the equation: \(q = mcΔT\), where m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. According to the conservation of energy principle, the heat gained by the cold water must be equal to the heat lost by the hot water. Thus, we have: \(m_{cold}c_{water}(T_{final} - T_{cold}) = m_{hot}c_{water}(T_{hot} - T_{final})\)

Step 2: Simplify the formula

Since we're working with water, the specific heat capacity, \(c_{water}\), is the same for both cold and hot water. Therefore, we can simplify the equation by dividing both sides by \(c_{water}\): \(m_{cold}(T_{final} - T_{cold}) = m_{hot}(T_{hot} - T_{final})\)

Step 3: Substitute the given values

Now, we can substitute the given values in the formula: • \(m_{cold} = 90.0 g\) • \(T_{cold} = 22.0^{\circ} C\) • \(T_{hot} = 55.0^{\circ} C\) • \(T_{final} = 37.0^{\circ} C\) \(90.0(37.0 - 22.0) = m_{hot}(55.0 - 37.0)\)

Step 4: Solve for m_hot

Solve the equation for \(m_{hot}\): \(90.0(15.0) = m_{hot}(18.0)\) \(1350 = 18m_{hot}\) Now, to find the mass of hot water, divide both sides by 18: \(m_{hot} = \frac{1350}{18}\) \(m_{hot} = 75.0 g\)

Step 5: State the final answer

The student needs to add 75.0 grams of hot water to the 90.0 grams of cold water to achieve a final temperature of 37.0°C.

Key Concepts

Specific Heat Capacity

Specific heat capacity is an important concept in chemistry and physics. It describes how much heat energy is required to raise the temperature of one unit of mass of a substance by one degree Celsius. For water, the specific heat capacity is quite high, at approximately 4.18 joules per gram per degree Celsius (J/g°C). This property means water can absorb a lot of heat without a significant increase in temperature. When performing calculations involving temperature changes, knowing the specific heat capacity allows you to determine how much heat energy will be needed or released.
In the context of this problem, it's vital to realize that the specific heat capacity for both hot and cold water is the same, simplifying our calculations. Whatever the temperature, the specific heat of water doesn't change significantly, making it possible to use the same value on either side of the equation.

Conservation of Energy

The conservation of energy is a fundamental principle stating that energy cannot be created or destroyed, only transformed or transferred. In the case of mixing hot and cold water, this translates to the energy (heat) lost by the hot water being equal to the energy gained by the cold water. Let's break it down:

• The cold water gains heat, increasing its temperature.
• The hot water loses heat, decreasing its temperature.

By applying this principle, the heat gained by the cold water should equal the heat lost by the hot water. We use the formula:
\[ m_{cold}c_{water}(T_{final} - T_{cold}) = m_{hot}c_{water}(T_{hot} - T_{final}) \]This equation shows that any gain in thermal energy for the cold water must be balanced by a corresponding loss for the hot water, ensuring that the overall energy within the system remains the same.

Temperature Change

Temperature change, or \(ΔT\), represents the difference between the initial and final temperatures of a substance. It is a crucial part of the specific heat equation:
\[ q = mcΔT \]In our exercise, \(ΔT\) helps us understand how much the temperature of the cold and hot water is changing. For the cold water initially at \(22.0^{\circ} C\), the temperature must rise to \(37.0^{\circ} C\). We calculate this change by subtracting the initial temperature from the final temperature:
\[ ΔT_{cold} = T_{final} - T_{cold} = 37.0^{\circ} C - 22.0^{\circ} C = 15.0^{\circ} C \]Similarly, for the hot water, the change is calculated by the difference from its original temperature down to the desired temperature:
\[ ΔT_{hot} = T_{hot} - T_{final} = 55.0^{\circ} C - 37.0^{\circ} C = 18.0^{\circ} C \]These calculations allow us to apply the conservation of energy principle effectively to find the unknown mass in this situation.

Mass Calculation

Mass calculation in this exercise involves finding the amount of hot water needed to reach a specific final temperature when mixed with a known mass of cold water.
The process begins by recognizing that the heat gained by the cold water equals the heat lost by the hot water. From the conservation of energy equation:
\[ m_{cold}(T_{final} - T_{cold}) = m_{hot}(T_{hot} - T_{final}) \]Substitute the known values into this equation:

• \(m_{cold} = 90.0 \text{ g}\)
• \(T_{final} - T_{cold} = 15.0^{\circ} C\)
• \(T_{hot} - T_{final} = 18.0^{\circ} C\)

Solve for \(m_{hot}\):
\[ 90.0 \times 15.0 = m_{hot} \times 18.0 \]\[ 1350 = 18 \times m_{hot} \]\[ m_{hot} = \frac{1350}{18} = 75.0 \text{ g} \]By performing these calculations, you determine the mass of hot water needed to achieve the desired final temperature, ensuring both portions of water reach a harmonious \(37.0^{\circ} C\).