If you can find a block of ice at 25ºC ... or even a picture or a folk legend of one ...
I'll go look up the conversion of joules to calories and work this out for you.
The temperature at which no more energy can be removed from a substance is absolute zero, which is 0 Kelvins or -273.15 degrees Celsius.
The boiling point of pure water is typically about 100 degrees Celsius. This value can vary based on factors such as pressure. Additionally, impurities in a sample of water can alter its boiling point.
0.385 Joules/Gram Celsius is the specific heat of copper. So, q(Joules) = mass * specific heat * change in temperature q = (200 g Cu)(0.385 J/gC)(30 C - 150 C) = - 9240 Joules -------------------------amount of heat dissipation ( answer can be positive )
Reduced
it is hot temperature but when thermal energy leaves the temperature is cool
The temperature at which no more energy can be removed from a substance is absolute zero, which is 0 Kelvins or -273.15 degrees Celsius.
6.276 kJ
6.276 kJ
-273.15 degrees Celsius (-459.67 degrees Fahrenheit) is the temperature at which no more energy can be removed from matter.It is called Absolute Zero and marks the 0 for the Kelvin and Rankine scale.
-273.15 degrees Celsius (-459.67 degrees Fahrenheit) is the temperature at which no more energy can be removed from matter.It is called Absolute Zero and marks the 0 for the Kelvin and Rankine scale.
There are no units for the temperature. The temperature change could be 230 Celsius degrees or 170 Fahrenheit degrees (94 Celsius degrees). In fact they could be in less commonly used units: eg Reaumur.
That sounds like a description of the temperature known as "absolute zero". This temperature is zero kelvin; it is also approximately -273 degrees Celsius.
This is because the ice at 0 deg C is colder to the extent that the latent heat of freezing has been removed from the water at 0 deg C.
The boiling point of pure water is typically about 100 degrees Celsius. This value can vary based on factors such as pressure. Additionally, impurities in a sample of water can alter its boiling point.
q( Joules) = mass grams * specific heat * change in temperature 90 kj = 90000 Joules 2 kg = 2000 grams 90000 Joules = (2000 g)( SP )(200 C - 80 C) 90000 = 240000(SP) 0.375 J/gC ( The actual specific heat of copper is 0.385 J/gC, so close enough for the problem type )
As of February 2013, the highest officially recorded temperature in Queensland was 49.5 degrees Celsius at Birdsville, in the state's southwest, on 24 December 1972.The highest unofficial temperature in Queensland was 53.1 degrees Celsius, recorded at Cloncurry in 1889. The outback Queensland town of Cloncurry originally held the record for the highest known temperature in the shade, at 53.1 °C (127.5 °F) on 16 January 1889. The Cloncurry record was later removed from Australian records because it was measured using unsuitable equipment (that is, not in a Stevenson screen, which only became widespread in Australian usage after about 1910).
E = mass x sp ht x Δ°tIn order to calculate the final temperature change, we must also find the temperature change.1. Find the temperature change.Divide joules by (mass x specific heat).Δ°t = 40,000J/500.0g x 4.184J/g°CΔ°t = 19.1°C2. Calculate the final temperature.Tf = 10.0 - 19.1Tf = -9.1°CThere was a decrease in temperature, indicating that it was an endothermic reaction (as energy was removed as stated in the question).