Heat

In physics and chemistry, heat is energy transferred from one body to another by thermal interactions.[1][2] The transfer of energy can occur in a variety of ways, among them conduction,[3] radiation,[4] and convection. Heat is not a property of a system or body, but instead is always associated with a process of some kind, and is synonymous with heat flow and heat transfer. Heat flow from hotter to colder systems occurs spontaneously, and is always accompanied by an increase in entropy. In a heat engine, internal energyof bodies is harnessed to provide useful work. The second law of thermodynamics prohibits heat flow directly from cold to hot systems, but with the aid of a heat pump external work can be used to transport internal energy indirectly from a cold to a hot body. Transfers of energy as heat are macroscopic processes. The origin and properties of heat can be understood through the statistical mechanics of microscopic constituents such as molecules and photons. For instance, heat flow can occur when the rapidly vibrating molecules in a high temperature body transfer some of their energy (by direct contact, radiation exchange, or other mechanisms) to the more slowly vibrating molecules in a lower temperature body. Example:

1:It takes 487.5 J to heat 25 grams of copper from 25 °C to 75 °C. What is the specific heat in Joules/g·°C? 2: What is the heat in Joules required to melt 25 grams of ice? What is the heat in calories?

Useful information: heat of fusion of water = 334 J/g = 80 cal/g

Solution:

1: Use the formula

q = mcΔT

where

q = heat energy

m = mass

c = specific heat

ΔT = change in temperature

487.5 J = (25 g)c(75 °C - 25 °C)

487.5 J = (25 g)c(50 °C)

Solve for c:

c = 487.5 J/(25g)(50 °C)

c = 0.39 J/g·°C

Answer:

The specific heat of copper is 0.39 J/g·°C.

2: Use the formula

q = m·ΔHf

where

q = heat energy

m = mass

ΔHf = heat of fusion

q = (25 g)x(334 J/g)

q = 8350 J

Part II

q = m·ΔHf

q = (25 g)x(80 cal/g)

q = 2000 ca

Answer:

The amount of heat required to melt 25 grams of ice is 8350 Joules or 2000 calories. Temperature scales

Much of the world uses the Celsius scale (°C) for most temperature measurements. It has the same incremental scaling as the Kelvinscale used by scientists, but fixes its null point, at 0°C = 273.15K, approximately the freezing point of water (at one atmosphere of pressure).[note 1] The United States uses the Fahrenheit scale for common purposes, a scale on which water freezes at 32 °F and boils at 212 °F (at one atmosphere of pressure). For practical purposes of scientific temperature measurement, the International System of Units (SI) defines a scale and unit for the thermodynamic temperature by using the easily reproducible temperature of the triple point of water as a second reference point. The reason for this choice is that, unlike the freezing and boiling point temperatures, the temperature at the triple point is independent of pressure (since the triple point is a fixed point on a two-dimensional plot of pressure vs. temperature). For historical reasons, the triple point temperature of water is fixed at 273.16 units of the measurement increment, which has been named the kelvin in honor of the Scottish physicist who first defined the scale. The unit symbol of the kelvin is K. Absolute zero is defined as a temperature of precisely 0 kelvins, which is equal to −273.15 °C or −459.67 °F.

The conversion of one temperature scale to another is sometimes required at nuclear facilities,and the operator should be acquainted with the process. The following two examples will behelpful.

Example 1: Temperature Scale ConversionWhat is the Rankine equivalent of 80°C? Solution:

°F= (9/5) °C + 32

= (9/5)(80) + 32

= 176 °F°R= °F + 460

= 176 + 460= 636 °R

Example 2:

Temperature Scale ConversionWhat is the Kelvin equivalent of 80°F? Solution:

°C= (5/9) (°F - 32)

= (5/9) (80 - 32)

=...