What is the heat energy required to raise the temperature called?

1) Calculate the energy required to boil 100ml of water for a cup of tea if the initial water temperature is 27.0°C. (The density of water is 1g/ml)

Since the density of water is 0.997g/ml , at 25oC we can round it to 1.00 g/mL. So 100ml of water has a mass of 100 grams. The change in temperature is (100°C - 27°C) = 73°C. Since the specific heat of water is 4.18J/g/°C we can calculate the amount of energy needed by the expression below.

Energy required = 4.18 J/g/°C X 100g X 73°C = 30.514KJ.

Try some exercises. 1) Calculate the energy needed to heat a) 80ml of water form 17°C to 50°C; b) 2.3 litres of water from 34°C to 100°C; c) 200g of cooking oil from 23°C to 100°C

Solutions

An astronaut in space needs to absorb 2,400KJ of solar energy in a container with an accurately known volume of water. The water's temperature is needed to increase from 20°C to 34.5°C. What amount of water is in the container?
Solution

3,450,560J of energy are absorbed by 300kg of water. If the initial temperature of the water is 20°C what is the final temperature?
Solution

A peanut of mass 2.34g is burnt in a calorimeter containing 100ml of water. If the temperature of the calorimeter rises from 23.5°C to 27.7°C calculate the energy content of the peanut in joules per gram.
Solution

When 0.15 gram of heptane C7H16 was burnt in a bomb calorimeter containing 1.5kg of water the temperature rose from 22.000°C to 23.155°C. Calculate the heat given out by heptane during combustion per mole. This is known as the heat of combustion.
Solution

Page 2

0.1 gram of methane was completely burnt in a bomb calorimeter containing 100ml of water. If the temperature increased by 11.82°C find the heat of combustion(energy released per mole)of methane.

The energy absorbed by the water is given by the expression below
Energy = heat capacity X temperature rise X mass of water
4.94KJ = 4.18J/g/°C X 11.82°C X 100
moles of methane = 0.1/16 = 0.00625mole
4.94 / 0.00625 = 790.4KJ/mole

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A brand of RAZMAN'S jelly beans was analysed for its energy content. A jelly bean of mass 1.5g was burnt completey in a bomb calorimeter containing 100ml of water. If the temperature rose from 21.56°C to 24.5°C calculate the energy content per gram of the jelly bean.

The energy absorbed by the water is given by the expression below
Energy = heat capacity X temperature rise X mass of water
1.23KJ = 4.18J/g/°C X 2.94°C X 100g
1.23/1.5 = 0.819KJ/g

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0.1 gram of carbon was burnt in a bomb calorimeter containing 200ml of water. If the temperature of the water increased by 3.92°C calculate the energy given off, during combustion, per mole of carbon.

The energy absorbed by the water is given by the expression below
Energy = heat capacity X temperature rise X mass of water
3277.12J = 4.18J/g/°C X 3.92°C X 200
Mole of carbon = 0.1 / 12 =0.0083

Energy per mole = 3277.12J / 0.0083 =394,819J/mole

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A 2.6 gram sample of sugar was burnt in a bomb calorimeter containing 100ml of water. The temperature of the water increased from 22.0°C to 24.3°C.

a)Calculate the amount of energy that was released in the burning of the sugar.

The energy absorbed by the water is given by the expression below
Energy = heat capacity X temperature rise X mass of water
961.4J = 4.18J/g/°C X 2.3°C X 100

b) Calculate the mole and the mass of sugar present.

Assuming all the energy released is absorbed by the water,every mole of sugar burnt must release 2803KJ of energy. The amount of sugar present in mole is given by the expression below 0.9614KJ/2803KJ = 0.00034mole

Mass of sugar = 0.00034 X 180 = 0.0612 grams

c) Assuming no other combustible material is present calculate the percent, by mass, of sugar in the sample.

(0.0612 / 2.6 ) X 100 = 2.35%

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The amount of heat energy required to raise the temperature of unit mass of a substance by 1 degree celsius is called

latent heat of vaporization

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The SI-unit of heat - or energy - is joule (J).

With temperature difference 

Other units used to quantify heat are the British Thermal Unit - Btu (the amount of heat to raise 1 lb of water by 1oF) and the Calorie (the amount of heat to raise 1 gram of water by 1oC (or 1 K)).

• more  about degrees Celsius and degrees Kelvin

A calorie is defined as the amount of heat required to change the temperature of one gram of liquid water by one degree Celsius (or one degree Kelvin).

1 cal = 4.184 J

1 J = 1 Ws

      = (1 Ws) (1/3600 h/s)

      = 2.78 10-4 Wh

      = 2.78 10-7 kWh

Heat Flow (Power)

Heat-transfer as result of temperature difference alone is referred to as heat flow. The SI units for heat flow is J/s or watt (W) - the same as power. One watt is defined as 1 J/s.

Specific Enthalpy

Specific Enthalpy is a measure of the total energy in a unit mass. The SI-unit commonly used is J/kg or kJ/kg.

The term relates to the total energy due to both pressure and temperature of a fluid (such as water or steam) at any given time and condition. More specifically enthalpy is the sum of internal energy and work done by applied pressure.

Heat Capacity

Heat Capacity of a system is

• the amount of heat required to change the temperature of the whole system by one degree.

Specific Heat

Specific heat  (= specific heat capacity) is the amount of heat required to change temperature of one mass unit of a substance by one degree.

Specific heat may be measured in J/g K, J/kg K, kJ/kg K, cal/gK or Btu/lboF and more. 

Never use tabulated values of heat capacity without checking the unites of the actual values!

• Specific heat unit converter

Specific heat for common products and materials can be found in the Material Properties section.

Specific Heat - Constant Pressure

The enthalpy - or internal energy -  of a substance is a function of its temperature and pressure.

The change in internal energy with respect to change in temperature at fixed pressure is the Specific Heat at constant pressure - cp.

Specific Heat - Constant Volume

The change in internal energy with respect to change in temperature at fixed volume is the Specific Heat at constant volume - cv.

Unless the pressure is extremely high the work done by applied pressure on solids and liquids can be neglected, and enthalpy can be represented by the internal energy component alone. Constant-volume and constant-pressure heats can be said to be equal.

For solids and liquids

cp = cv                                            (1)

The specific heat represents the amount of energy required to raise 1 kg of substance by 1oC (or 1 K), and can be thought of as the ability to absorb heat. The SI units of specific heats are J/kgK (kJ/kgoC). Water has a large specific heat of 4.19 kJ/kgoC compared to many other fluids and materials.

• Water is a good heat carrier!

Amount of Heat Required to Rise Temperature

The amount of heat needed to heat a subject from one temperature level to an other can be expressed as:

Q = cp m dT                                                (2)

where

Q = amount of heat (kJ)

cp = specific heat (kJ/kgK)

m = mass (kg)

dT = temperature difference between hot and cold side (K)

Example Heating Water

Consider the energy required to heat 1.0 kg of water from 0 oC to 100 oC when the specific heat of water is 4.19 kJ/kgoC:

Q = (4.19 kJ/kgoC) (1.0 kg) ((100 oC) - (0 oC))

    = 419 (kJ)

Work

Work and energy are from a technical viewpoint the same entity - but work is the result when a directional force (vector) moves an object in the same direction.

The amount of mechanical work done can be determined by an equation derived from Newtonian mechanics

Work = Applied force x Distance moved in the direction of the force 

or 

W = F l                                              (3)

where 

W = work (Nm, J)

F = applied force (N)

l = length or distance moved (m)

Work can also be described as the product of the applied pressure and the displaced volume:

Work = Applied pressure x Displaced volume

or

W = p A l                                             (3b)

where

p = applied pressure (N/m2, Pa)

A = pressurized area (m2)

l = length or distance the pressurized area is moved by the applied force (m)

Example - Work done by a Force

The work done by a force 100 N moving a body 50 m can be calculated as 

W = (100 N) (50 m)

  = 5000 (Nm, J)

The unit of work is joule, J, which is defined as the amount of work done when a force of 1 newton acts for a distance of 1 m in the direction of the force.

1 J = 1 Nm

Example - Work due to Gravitational Force

The work done when lifting a mass of 100 kg an elevation of 10 m can be calculated as 

W = Fg h

   = m g h

  = (100 kg) (9.81 m/s2) (10 m)

  = 9810 (Nm, J)

where

Fg = force of gravity - or weight (N)

g = acceleration of gravity 9.81 (m/s2)

h = elevation (m)

In imperial units a unit work is done when a weight of 1 lbf (pound-force) is lifted vertically against gravity through a distance of 1 foot. The unit is called lb ft.

An object with mass 10 slugs is lifted 10 feet. The work done can be calculated as

  W = Fg h

     = m g h

     = (10 slugs) (32.17405 ft/s2) (10 feet)

     = 3217 lbf ft

Example - Work due to Change in Velocity

The work done when a mass of 100 kg is accelerated from a velocity of 10 m/s to a velocity of 20 m/s can be calculated as

W = (v22 - v12) m / 2

  = ((20 m/s)2 - (10 m/s)2) (100 kg) / 2

  = 15000 (Nm, J)

where

v2 = final velocity (m/s)

v1 = initial velocity (m/s)

Energy

Energy is the capacity to do work (a translation from Greek-"work within"). The SI unit for work and energy is the joule, defined as 1 Nm.

Moving objects can do work because they have kinetic energy. ("kinetic" means "motion" in Greek).

The amount of kinetic energy possessed by an object can be calculated as

Ek =1/2 m v2                                             (4)  

where

m = mass of the object (kg)

v = velocity (m/s)

The energy of a level position (stored energy) is called potential energy. This is energy associated with forces of attraction and repulsion between objects (gravity).

The total energy of a system is composed of the internal, potential and kinetic energy. The temperature of a substance is directly related to its internal energy. The internal energy is associated with the motion, interaction and bonding of the molecules within a substance. The external energy of a substance is associated with its velocity and location, and is the sum of its potential and kinetic energy.