/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 You have \(750 \mathrm{~g}\) of ... [FREE SOLUTION] | 91Ó°ÊÓ

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You have \(750 \mathrm{~g}\) of water at \(10.0^{\circ} \mathrm{C}\) in a large insulated beaker. How much boiling water at \(100.0^{\circ} \mathrm{C}\) must you add to this beaker so that the final temperature of the mixture will be \(75^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The mass of boiling water required can be calculated using the formulas from step 1 and step 2. The final answer depends on the values calculated.

Step by step solution

01

Determine the heat gained by the cooler water

Using the formula \(Q = mc\Delta T\), where Q is the heat energy, m is the mass, c is the specific heat capacity of water, and \(\Delta T\) is the change in temperature, find the heat gained by the cooler water. The specific heat capacity of water is approximately \(4.18 \mathrm{J/g^{\circ}C}\). Subtract the initial temperature from the final temperature to get \(\Delta T\). The cooler water's mass is 750g, so the formula becomes \(Q = 750g * 4.18 \mathrm{J/g^{\circ}C} * (75^{\circ}C - 10^{\circ}C)\).
02

Determine the mass of the boiling water

The heat gained by the cooler water is the same as the heat lost by the hotter one. Use the same formula, but solve for m instead. This means that \(Q = mc\Delta T\) becomes \(m = Q / (c * \Delta T)\). In this case the \(\Delta T\) is the difference between the boiling water's initial temperature and the final mixture temperature, so the formula becomes \(m = Q / (4.18 \mathrm{J/g^{\circ}C}* (100^{\circ}C - 75^{\circ}C)\).
03

Evaluate the values

First calculate the value for Q from step 1 and then substitute this value for Q in the formula from step 2 to find the mass of the boiling water needed to raise the temperature of the cooler water.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat Capacity
The measure of the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin) is known as specific heat capacity (often simplified as 'specific heat'). It's a fundamental property of materials and plays a crucial role in the process of heating and cooling.

In a more technical sense, specific heat capacity, represented by the symbol c, is described by the equation: \[\begin{equation} Q = mc\triangle T \text{where} \begin{itemize} \tn\text{Q is the heat energy in joules (J)}, \tm is the mass of the object in grams (g) or kilograms (kg),\text{c is the specific heat capacity in J/g}\textdegree\text{C} or J/kg}\textdegree\text{C}, \text{and \(\triangle T\) is the change in temperature in Celsius (}\textdegree\text{C}). \end{itemize} \text{}\end{equation}\] To give an example, water has a specific heat capacity of approximately 4.18 J/g}\textdegree}{C}\text{, which is relatively high compared to other substances. This means that water changes its temperature slowly with the addition or removal of heat, making it an excellent medium for thermal regulation.

Understanding specific heat capacity is crucial when attempting to alter the temperature of any object or substance, whether you're heating up dinner or performing calculations in a physics problem. It explains why different substances heat up or cool down at different rates when absorbing or losing the same amount of heat energy.
Heat Transfer
Heat transfer refers to the movement of heat energy from one object or substance to another, or from one part of an object to another, due to a temperature difference. It is a foundational concept in thermodynamics and can occur in three main ways: conduction, convection, and radiation.

Conduction is the transfer of heat through a solid material, convection is the transfer of heat by the movement of fluids (liquids or gases), and radiation is the transfer of heat in the form of electromagnetic waves. In the context of our textbook example, we are primarily concerned with the transfer of heat through conduction and convection in the water.

When two bodies of water at different temperatures are mixed, as in the given problem, heat will transfer from the hotter to the cooler one until thermal equilibrium is reached. The final temperature of the mixture will be somewhere between the initial temperatures of the individual quantities, depending on their masses and specific heat capacities. The calculation of the final temperature, or the amount of water required to reach a specific temperature, involves understanding and applying the principles of heat transfer.
Temperature Change Calculation
Temperature change calculation is the process used to determine the change in temperature that a substance undergoes when heat is added or removed. This change is called \[\begin{equation}\triangle T \text{or the difference between the final (}\text{T}_\text{f}\text{) and initial (}\text{T}_\text{i}\text{) temperatures}: \triangle T = \text{T}_\text{f} - \text{T}_\text{i}\text{}\end{equation}\]textA key application of this calculation is in solving problems like the one provided from the textbook, where we must find the amount of boiling water needed to achieve a desired final temperature when mixed with cooler water.

The formula employed in this case combines the concepts of specific heat capacity and heat transfer: \[\begin{equation}Q = mc\triangle T\text{}\end{equation}\]This equation allows us to calculate the heat energy required (\text{Q}), which is equal to the energy lost or gained by the water. By rearranging the equation, we can solve for any unknown variable, given that the others are known. This process is invaluable in a variety of practical situations, such as cooking, climate control, and even complex industrial processes.

For students, learning to calculate temperature changes not only assists in completing textbook exercises, but it also provides a fundamental understanding of thermal processes that underpin various scientific and engineering fields.

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Most popular questions from this chapter

If the air temperature is the same as the temperature of your skin (about \(30^{\circ} \mathrm{C}\) ), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical \(70 \mathrm{~kg}\) person's body produces energy at a rate of about \(500 \mathrm{~W}\) due to metabolism, \(80 \%\) of which is converted to heat. (a) How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) (b) The evaporated water must, of course, be replenished, or the person will dehydrate. How many \(750 \mathrm{~mL}\) bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is \(1.0 \mathrm{~kg} .\) )

In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a \(200 \mathrm{~W}\) electric immersion heater in \(0.320 \mathrm{~kg}\) of water. (a) How much heat must be added to the water to raise its temperature from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C} ?\) (b) How much time is required? Assume that all of the heater's power goes into heating the water.

\(Camels require very little }}\) water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a \(400 \mathrm{~kg}\) camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day ( 12 hours) instead of letting it rise to \(40.0^{\circ} \mathrm{C}\). (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) )

An insulated beaker with negligible mass contains \(0.250 \mathrm{~kg}\) of water at \(75.0^{\circ} \mathrm{C}\). How many kilograms of ice at \(-20.0^{\circ} \mathrm{C}\) must be dropped into the water to make the final temperature of the system \(40.0^{\circ} \mathrm{C}\) ?

A \(\mathrm 500.0 \mathrm{~g}\) chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing \(1.00 \mathrm{~kg}\) of water at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) After waiting and gently stirring for 5.00 minutes, you observe that the water's temperature has reached a constant value of \(22.0^{\circ} \mathrm{C}\). (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? (b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain. (c) If the heat absorbed by the Styrofoam actually is not negligible, how would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.

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