167.2J if it doesn't have to do with phase changes.
Use this formula. q(in Joules) = Mass * specific heat * change in temperature I will use specific heat of water at 25 C. You can look up specific heat of steam. You say " heat to " so I assume you have final and initial heat backwards. q = 25.0 grams H2O * 4.180 J/gC * (100.0 C - 29.25 C ) q = 7393 Joules
I assume you mean 30o Celsius. Use this formula.q(joules) = mass * specific heat * change in temperatureq = (15 grams water)(4.180 J/gC)(40o C - 30o C)= 627 joules==========( perhaps 630 joules to be in significant figures territory )
400 joules.
the formula for heat is: joules= change in temp* mass* cpby: Professor Thompson
If all 1700 Joules of work get converted into heat, then, of course, you get 1700 Joules of heat.
134 joules. You're very welcome for answering your question.
The needed heat is 47,65 Joules.
The necessary heat is 9,22 joules.
20 degree c\
It takes 4.186 Joules to heat one gram of water by 1-degree Celsius. 4.186 * 4000 = 16,744 Joules to heat 4 kilos of water by 1-degree. 16,744 * 70 = 1,172,080 Joules. The above assumes that one litre of water weighs exactly 1 Kilogram.
0.0796
q(joules) = mass * specific heat * change in temperatureq = (55.0 g H2O)(4.180 J/gC)(25.5o C - 60o C)= (- ) 7932 Joules=============heat change
The answer is 55,117 kJ.
The amount of water whose temperature would change by 15 degrees Celsius when it absorbs 2646 joules of heat energy is 42,2g H2O.
10ml's of water is equal to 10cm3 of water. 10cm3 of water has a mass of 10g. The specific heat of water is 4.134 J/K. The change in temperature is 1 degree Kelvin. Use Q=mC∆T which means Heat= (Mass)(Specific Heat)(Change in Temperature) Q= (10)(4.134)(1) Q=(10)(4.134) Q=41.34 Joules
334.8 Joules
10 degrees Celsius