Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Experience the ease of finding accurate answers to your questions from a knowledgeable community of professionals. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
Qt = 4.719kJ heat, is required to warm 10.0 g of ice, initially at -10.0 degrees Celsius, to steam at 110 degrees celsius. The heat capacity of ice is, c = 2.09 J/g C.
The mass of ice is, m = 10.0 g.
The initial temperature is, T = - 10 degree Celsius.
The final temperature is, T' = 110.0 degree Celsius.
The heat capacity of ice is, c = 2.09 J/g C.
The heat capacity of steam is, c' = 2.01 J/g C.
When ice absorbs heat, the conversion takes place from ice to water (liquid) and from the liquid to steam at 110 degree Celsius. So,
The amount of heat absorbed is given as,
Q = m[c(T' - T) + c'T']
Q = 10[2.09(110-(-10)+(2.01*110)]
Q = 4719J
Convert in kJ as,
Q = 4719/1000
Q = 4.719 kJ
Thus, we can conclude that the amount of heat required to warm the ice is 4.719 kJ.
Learn more about steam here
https://brainly.com/question/15447025
#SPJ4
The complete question is -
How much heat (in kJ) is required to warm 10.0 g of ice, initially at −10.0 °C, to steam at 110.0 °C? The heat capacity of ice is 2.09 J/g · °C, and that of steam is 2.01 J/g °C.
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
