Q.

Heat capacity of water is 18 cal-degree-1-mol-1. The quantity of heat needed to rise temperature of 18g water by 0.2oC is X cal. Then amount of CH4(g) to be burnt to produce X cal heat is(CH4+2O2 →CO2+2H2O, ΔH= –200K.Cal)

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

answer is 4.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

q =m CpΔT     = 3.6 cal16gm -------200Kcal?      -------3.6 x 10-3 Kcal    = 288 x 10-6gm= 0.288 mg
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
whats app icon
Heat capacity of water is 18 cal-degree-1-mol-1. The quantity of heat needed to rise temperature of 18g water by 0.2oC is X cal. Then amount of CH4(g) to be burnt to produce X cal heat is(CH4+2O2 →CO2+2H2O, ΔH= –200K.Cal)