40.5556 C
Q=6*550*1.00q=3300
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.
To determine how many degrees J will raise the temperature of g of water, we need to use the specific heat capacity formula: ( Q = mc\Delta T ), where ( Q ) is the heat added (in joules), ( m ) is the mass of the water (in grams), ( c ) is the specific heat capacity of water (approximately 4.18 J/g°C), and ( \Delta T ) is the change in temperature (in °C). Rearranging the formula gives ( \Delta T = \frac{Q}{mc} ). Without specific values for Q and g, we cannot calculate the exact change in temperature.
0.11cal/g degrees C
There are 4 main differences between C programming and VHDL programming. C is a mid-level language, while VHDL is a hardware description language. C can handle one type of instruction, while VHDL can handle two. C does not require as much resource usage as VHDL. C can be written only with logical thinking, but a VHDL programmer must understand hardware circuits.
To compare the temperatures, first, convert the inside temperature from Fahrenheit to Celsius. Using the formula (C = \frac{5}{9}(F - 32)), the inside temperature of 75°F converts to approximately 24°C. Therefore, the outside temperature of 28°C is about 4°C warmer than the inside temperature.
about -64*C
32° F = Zero °C .
It is 14.44... (repeating) deg C.
To convert Celsius to Kelvin, add 273.15 to the Celsius temperature. Therefore, if it is 65 degrees Celsius outside, it would be 338.15 Kelvin.
According to my thermometer, as i have measured it, on July 18,2009 - 6:20am --- the temperature outside of my house was 22'C+, then when i travel, i put the thermometer outside of the car, then the temperature was.. 19'C - 20.5'C..... :)
37 c
Unfortunately, most domestic A/C units are designed to be set at 20 degrees below the outside ambient temperature. Setting it lower will overstress the system. So if it's 100 degrees outside, then your A/C should be set to 80 degrees to avoid breakdown.
Diastase works best at a temperature range of 40-50°C (104-122°F). Temperatures outside this range can denature the enzyme, reducing its effectiveness.
USUALLY YOUR A/C SYSTEM IS MALFUNCTIONING
30 degrees F because 30 degrees C is about 90 degrees F and you would not have much ice on the lake!
C = 5/9 (f-32) = 5/9 (85-32) = 5/9 (53) = 29.4c