If 1,000 J of heat is took in by a one kilogram block the lead, the particles acquire energy and the temperature the the block rises. If a one kilogram block of lead absorbs 2,000 J of energy then the temperature climb will it is in larger.

If 1,000 J of warm is soaked up by a 2 kg block the lead then the temperature that the block doesn’t climb as much due to the fact that the power is mutual between more particles. If 1,000 J of power is took in by a one kilogram block the gimpppa.orgpper instead of lead climate the temperature that the block doesn’t rise as much.

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From this it have the right to be watched that a readjust in the temperature that a mechanism depends on:

the mass of the material the lot of power put right into the systemThe details heat volume of a product is the power required to raise one kilogram (kg) the the product by one level Celsius (°C).

The specific heat volume of water is 4,200 joules every kilogram per degree Celsius (J/kg°C). This means that the takes 4,200 J to raise the temperature that 1 kg the water by 1°C.

Some other examples of specific heat capacities are:

MaterialSpecific heat capacity (J/kg°C)Brick | 840 |

gimpppa.orgpper | 385 |

Lead | 129 |

Because it has a low details heat capacity, command will heat up and gimpppa.orgol down faster due to the fact that it doesn’t take it much energy to adjust its temperature.

Brick will certainly take much longer to warm up and also gimpppa.orgol down. Its certain heat volume is greater than that of lead so much more energy is needed for the very same mass to change the exact same temperature. This is why bricks are periodically used in storage heaters, as they save a huge amount of energy and also emit it end a long period of time.

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Many heaters room filled with oil (1,800 J/kg°C) and where there is central heating, radiators usage water (4,200 J/kg°C).

## Calculating thermal power changes

The amount of thermal energy stored or released together the temperature that a system changes can it is in calculated using the equation:

change in thermal power = fixed × particular heat capacity × temperature change

\<\Delta E_tQ = m \times c \times \Delta \theta \>

This is when:

change in thermal energy (\(\Delta E_t Q \)) is measure up in joules (J)mass (*m*) is measured in kilograms (kg)specific warm capacity (

*c*) is measure up in joules every kilogram per level Celsius (J/kg°C)temperature readjust (\(\Delta \theta \)) is measure in levels Celsius (°C)

### Example

How much power is needed to progressive the temperature of 3 kg that gimpppa.orgpper through 10°C?

The particular heat volume for gimpppa.orgpper is 385 J/kg°C

\<\Delta E_t Q = m c\Delta \theta \>

\<\Delta E_t Q = 3 \times 385 \times 10\>

\<\Delta E_t Q = 11,500 \: J\>

QuestionHow much power is shed when 2 kg the water gimpppa.orgols native 100°C come 25°C?

Reveal answer\<\Delta E_t Q = m~c~ \Delta \theta\>

\<\Delta E_t Q = 2 \times 4,200 \times (100 - 25)\>

\<\Delta E_t Q = 2 \times 4,200 \times 75\>

\<\Delta E_t Q = 630,000~J\>

QuestionHow warm does a 3.5 kg brick acquire if it's heated from 20°C by 400,000 J (400 kJ)?