Check the duo-valve first, then the heater register
Eight degrees below zero Celsius. It is a subtraction problem.
The freezing temperate in Fahrenheit is exactly 32 degrees. But as you may know that temperature is different in Celsius. So the freezing temperature in Celsius is different that Fahrenheit. Like I said before the freezing temperature in Fahrenheit is 32 degrees. But as I have been saying, that temperature is different in Celsius. I do know that the difference between the temperature in Fahrenheit and the temperature in Celsius is 32 degrees. So if you do the mathematical problem (32-32) correctly you will get the answer to your question which is "Freezing point in Celsius temp?". Therefor I am clearly stating that the freezing temperature in Celsius is 0 degrees.
This is a mathematical word problem. Since we're using the modifier "decreases" we understand our problem will be substraction.-60C - 80C = -140C.
Here is a fine temperature converter. Scroll down to related links and look at "Conversion of temperature units".
Celsius or Fahrenheit? If Celsius, then definitely a problem! If Fahrenheit, turn your device off for a little bit because you have been working it to hard.
Let C be the temperature reading in degrees Celsius and F be the temperature reading in degrees Fahrenheit. Formula: C = (F - 32) * 5 / 9, or its equivalent, F = (C * 9 / 5) + 32. Use the second expression for the problem. F = ( 7 * 9 / 5) + 32 = 44.6. An aside: To check your formula, in case it is not given, plug in C = 0 and you should get F = 32. Plug in C = -40 and you should get F = -40 too! This second fact is a popular math problem. ==========================
327"Zero" in kelvin is absolute zero, which is -273.15o Celsius. In order to convert from Kelvin to Celsius (From Ko to Co) we subtract absolute zero from our degrees in kelvin or...Co = Ko- 273.15This is because the interval from one degree to another is the same as Celsius, just shifted due to the change in what we define as "zero."To answer your specific problem:Co = Ko- 273.15Co = 600- 273.15Co = 326.85 The answer is B:327 degrees C i just took the test
Two q's to 0 problem. q(Joules) = mass * specific heat * change in temperature (50.0 g H2O)(4.180 J/gC)(Tf - 10.0 C) + (10.0 g H2O)(4.180 J/gC)(Tf - 50.0 C) = 0 209Tf - 2090 + 41.8Tf - 2090 = 0 250.8Tf = 4180 Temperature Final = 16.7 Celsius --------------------------------------------
The relevant equation behind this problem is Q=m*c* ΔT Where Q is the energy that must be added to or taken from the system, m is the mass of the object, c is the objects specific heat, and ΔT is the change in temperature in Celsius or Kelvin. Plugging in the given values we get that Q=.015kg * 128J/(kg*C) * 10C=19.2J. Therefore, you need 19.2 joules of heat in order to raise the temperature of a .015kg sample of lead by 10 degrees Celsius.
It is: 5/9*(-391-32) = -235 degrees Celsius
-220 degrees Celsius = -364 degrees FahrenheitThe conversion formula is [Celsius] = [Fahrenheit-32] * 5/9. A quick way to check if the formula is correctly reproduced from memory is to try to plug (-40) into either Celsius or Fahrenheit. Try it -- it is a surprise.Ans for the problem = (-220)*9/5 + 32 = -364.
Let F and C be the temperature in degrees Fahrenheit and Celsius, respectively. Known: F = (C*9/5) + 32 or C = (F-32)*5/9 In this problem, C = (40 - 32)*5/9 = 4.44... ================
One of the problems with the Celsius and Fahrenheit scales is that they are not linear. We cannot say, for example, that a cup of water at 40 degrees C is twice as hot as one as 20 degrees, or that water at 20 degrees is twice as hot as water at 10 degrees. The absolute -- or Kelvin -- scale solves this problem, because it is linear.
Sorry but that it way to simple for me to give you the answer for. The way you solve it is you find the difference between the numbers. In other words, you subtract the two numbers. The problem looks like this: -38--68 or -68--38. If you are having trouble with this problem, I would suggest going to your teacher for some extra help. The good teachers are usually more then willing to help struggling students.
37 degrees Centigrade is the same as 98.6 degrees Fahrenheit, which is normal body temperature. 38 degrees is equal to 100.4, so it shouldn't be a problem
For this problem, we are looking at two variables: pressure and temperature. Therefore, we will use Gay-Lussac's law orP1 ÷ T1 = P2 ÷ P2P1 = 3.00 ATMT1 = 2.00oC = 275 K*P2 = PT2 = 10.0oC = 283 K**Note that all temperatures must be in Kelvin3.00/275 = P/283P2 = 3.09 ATM
11Since the morning temperature was -3 degrees & the temperature had risen 14 degrees by 3pm, the problem is simply done by adding -3 & 14 which is 11.
This is just a division problem, where Q is the change in energy T is the change in temperature in degrees C (or K) C is the specific heat m is the mass C = ΔQ/ΔT and Q = CmT 1.33 joules = (0.129 joule/gram degree)(5 grams)(T degrees) 1.33 = 0.645 T T = 1.33 / 0.645 = 2.06 degrees Celsius
There is not enough information in this problem to answer the question. You must know the mass of the iron to find the amount of joules used to heat the iron.
20 degrees C of temperature is often comfortable for most people. It is not very hot or cold and most people can easily survive this temperature without much of a problem.
You mean how much heat energy will be lost/transferred as you are losing Joules here. All in steam, so a simple q problem and no change of state. 2.67 kg = 2670 grams q = (2670 grams steam)(2.0 J/gC)(105 C - 282 C) = - 9.45 X 105 Joules ----------------------------------- This much heat energy must be lost to lower the temperature of the steam.
Let T = temperature T is less than 40 degrees. T < 40 degrees
Based on this page, which contains a slightly more complex problem the answer would be 40 degrees, as (60*100+20*100)/(100+100) =40. http://www.crystalgrowing.com/recipes/solution_calculator/solution_calculator.htm
You can't. Celsius per mmHg is a relationship of temperature to pressure. You can however solve for temperature if you have the value of pressure (e.g. if P= 10 mmHg and V/P = 2 ºC/mmHgthen V= (2 ºC/mmHg)(10 mmHg) = 20 ºC). If you have more information in the problem you might be referring to Gay-Lussac's Law, which compares two values of pressure and temperature to show the relationship (GL'sL: P1/T1 = P2/T2 ... but that would be pressure per unit volume). I'm not sure what you are looking for exactly, but you can't convert temp. to pressure (just like you can't convert feet to lbs.).