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One way to keep the contents of a garage from becoming too cold on a night when a severe subfreezing temperature is forecast is to put a tub of water in the garage. If the mass of the water is \(125 \mathrm{~kg}\) and its initial temperature is \(20^{\circ} \mathrm{C},\) (a) how much energy must the water transfer to its surroundings in order to freeze completely and (b) what is the lowest possible temperature of the water and its surroundings until that happens?

Short Answer

Expert verified
The water must transfer 52,215,000 J of energy in total. The lowest temperature is 0°C.

Step by step solution

01

Understanding the Problem

We need to calculate how much energy the water must transfer to freeze completely and what the minimum temperature can be.
02

Define Required Constants

The specific heat capacity of water is approximately \(4.186 \, \text{J/g°C}\), the latent heat of fusion is \(334,000 \, \text{J/kg}\), and water freezes at \(0^{\circ} \text{C}\).
03

Calculate Energy to Cool to 0°C

First, we calculate the energy required to lower the water's temperature from \(20^{\circ} C\) to \(0^{\circ} C\). This is done using the formula: \( Q = mc\Delta T \), where \( m = 125,000 \text{ g} \), \( c = 4.186 \, \text{J/g°C} \), and \( \Delta T = 20°C \).
04

Calculation 1 - Energy to Cool

\[ Q_1 = 125,000 \, \text{g} \times 4.186 \, \text{J/g°C} \times 20°C = 10,465,000 \, \text{J} \]
05

Calculate Energy to Freeze Water

Next, calculate the energy required to freeze the water using the latent heat of fusion: \( Q = mL \), where \( m = 125 \, \text{kg} \) and \( L = 334,000 \, \text{J/kg} \).
06

Calculation 2 - Energy to Freeze

\[ Q_2 = 125 \, \text{kg} \times 334,000 \, \text{J/kg} = 41,750,000 \, \text{J} \]
07

Total Energy Transfer

The total energy transferred by the water to its surroundings is the sum of cooling and freezing: \( Q_{\text{total}} = Q_1 + Q_2 \).
08

Total Calculation

\[ Q_{\text{total}} = 10,465,000 \, \text{J} + 41,750,000 \, \text{J} = 52,215,000 \, \text{J} \]
09

Lowest Possible Temperature

The lowest possible temperature for the water and its surroundings before it completely freezes is \(0^{\circ} \text{C}\), as water begins to transition to ice at this temperature.

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Key Concepts

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

Understanding Specific Heat Capacity
Specific heat capacity is a fundamental concept in thermodynamics that describes how much energy a substance can absorb or release as its temperature changes. It's the amount of heat energy required to change the temperature of one gram of a substance by one degree Celsius. For water, this value is approximately \(4.186 \, \text{J/g°C}\).

This property makes water an excellent medium for thermal regulation, such as preventing rapid temperature drops in the garage environment by absorbing heat. When water at \(20^{\circ} \mathrm{C}\) cools to \(0^{\circ} \mathrm{C}\), the energy transferred can be calculated with the formula: \[Q = mc\Delta T\]

Where:
  • \(m\) is the mass, in grams, of the water which in this case is \(125,000 \, \text{g}\).
  • \(c\) is the specific heat capacity of water \(4.186 \, \text{J/g°C}\).
  • \(\Delta T\) is the temperature change \(20°C \to 0°C\).
This results in a calculated value of \(10,465,000 \, \text{J}\) being released as the water temperature drops to just freezing.
Exploring Latent Heat of Fusion
The latent heat of fusion is the energy required to change a substance from solid to liquid (or vice versa) at a constant temperature. For water, this is \(334,000 \, \text{J/kg}\).

This property is crucial in the case of the water's transition from liquid to solid as it involves energy transfer without a temperature change. Here's the reasoning:
  • When the water reaches \(0^{\circ} \text{C}\), it must release additional energy for the phase change from liquid to ice.
  • This is calculated with \(Q = mL\), where \(m\) is mass and \(L\) is the latent heat of fusion.
  • For the given problem, the energy required for the complete phase change is \(41,750,000 \, \text{J}\).
It means even when the temperature is held constant at freezing, considerable energy is still released, important for keeping the garage above freezing temperature for an extended period.
Breakdown of Energy Transfer
Energy transfer in thermodynamics, especially dealing with water, is about the exchange of energy due to both temperature change and phase transitions. Combined, these calculations give us insights into how much energy the water needs to expel to freeze.

When adding the energy needed to cool the water and then freeze it, the total energy transferred to the surroundings is:
  • From cooling: \(10,465,000 \, \text{J}\)
  • From freezing: \(41,750,000 \, \text{J}\)
So, the entire energy transferred is \(52,215,000 \, \text{J}\).

This comprehensive energy exchange keeps the garage from dipping too low in temperature. The lowest possible temperature that can be achieved is maintained at \(0^{\circ} \text{C}\) until all the water is frozen, making it an effective temperature buffer.

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Most popular questions from this chapter

A lab sample of gas is taken through cycle abca shown in the \(p-V\) diagram of Fig. 18 -43. The net work done is \(+1.2 \mathrm{~J}\). Along path \(a b,\) the change in the internal energy is \(+3.0 \mathrm{~J}\) and the magnitude of the work done is \(5.0 \mathrm{~J} .\) Along path \(c a,\) the energy transferred to the gas as heat is \(+2.5 \mathrm{~J}\). How much energy is transferred as heat along (a) path \(a b\) and (b) path \(b c ?\)

A rectangular plate of glass initially has the dimensions \(0.200 \mathrm{~m}\) by \(0.300 \mathrm{~m}\). The coefficient of linear expansion for the glass is \(9.00 \times 10^{-6} / \mathrm{K} .\) What is the change in the plate's area if its temperature is increased by \(20.0 \mathrm{~K} ?\)

An aluminum-alloy rod has a length of \(10.000 \mathrm{~cm}\) at \(20.000^{\circ} \mathrm{C}\) and a length of \(10.015 \mathrm{~cm}\) at the boiling point of water. (a) What is the length of the rod at the freezing point of water? (b) What is the temperature if the length of the rod is \(10.009 \mathrm{~cm} ?\)

Nonmetric version: (a) How long does a \(2.0 \times 10^{5}\) Btu/h water heater take to raise the temperature of 40 gal of water from \(70^{\circ} \mathrm{F}\) to \(100^{\circ} \mathrm{F}\) ? Metric version: (b) How long does a \(59 \mathrm{~kW}\) water heater take to raise the temperature of \(150 \mathrm{~L}\) of water from \(21^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\) ?

A vertical glass tube of length \(L=1.280000 \mathrm{~m}\) is half filled with a liquid at \(20.000000^{\circ} \mathrm{C}\). How much will the height of the liquid column change when the tube and liquid are heated to \(30.000000^{\circ} \mathrm{C} ?\) Use coefficients \(\alpha_{\text {glass }}=1.000000 \times 10^{-5} / \mathrm{K}\) and \(\beta_{\text {liquid }}=4.000000 \times 10^{-5} / \mathrm{K}\)

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