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∙ 13y agoBy definition one calory is needed to heat up one gram of water exactly one degree celsius.
So 100 cal will heat 10 g. of water up for 10oC:
Temperature after heating will be 15 oC + 10oC = 25 oC
Wiki User
∙ 13y agoThe temperature of the water would reach 25°C. The specific heat capacity of water is 4.18 J/g°C, so to calculate the temperature change, we can use the formula: Q = mcΔT, where Q is the heat absorbed, m is the mass of water, c is the specific heat capacity of water, and ΔT is the temperature change. By rearranging the formula to solve for ΔT, we find that the temperature change is 10°C. This means the final temperature would be 25°C (15°C + 10°C).
The temperature drop from 3 degrees Celsius to -7 degrees Celsius is 10 degrees.
The dew point temperature is 21 degrees Celsius when the dry bulb temperature is 24 degrees Celsius and the wet bulb temperature is 22 degrees Celsius.
The new temperature would be -5 degrees Celsius.
To measure temperature in metric units, you would use Celsius degrees. Fahrenheit degrees are not considered metric units of temperature measurement.
The temperature on Ceres varies from -38 Celsius (highest), -105 Celsius (average), -114 Celsius (lowest).
To transform 1 gram of ice at 0 degrees Celsius to 1 gram of water vapor at 100 degrees Celsius, 720 calories are added (absorbed). There are no calories released during the process.
False. Heat is measured in units of energy such as calories or joules, not in degrees Celsius. Temperature, on the other hand, is measured in degrees Celsius.
Change in temp = 28° - 22° = 6°. Heat capacity of liquid water is 1 calorie per gram per degree(C).So the energy absorbed is (50) x (6) = 300 calories.
If the temperature decreases by 30 degrees Celsius from 20 degrees Celsius, the new temperature will be -10 degrees Celsius.
Liquid water absorbs 1 calorie per (gram °C)Temperature difference = 100°C - 23°C = 77°C(50 g) * (77°C) * (1 cal/g/°C) = 3850 calories
A calorie is the amount of heat you need to raise the temperature of one gram of water by one degree Celsius. Assuming you are raising the temperature of the water from twenty degrees Celsius to ninety-nine degrees Celsius, it would take 20,000 calories. To calculate this, subtract 20 from 99. This is the amount of degrees you need to raise the temperature of the water by. Then multiply that number by 256, the amount of water in grams. You should get 20,244 calories. In significant digits, your answer should be 20,000 calories.
105C
The temperature drop from 3 degrees Celsius to -7 degrees Celsius is 10 degrees.
35 degrees Celsius.
The temperature is 0 degrees Celsius.
Zero degrees Celsius.
The temperature of ice is 0 degrees Celsius. At this temperature, ice is in its solid form.