Question

how much energy is transferred when 1 gram of boiling water at 100 degree Celsius cools to water at 0 degree celsius

Answer 1

mass of water (m) = 1g
initial temperature (T1) = 0 c
Final temperature (T2) = 100 c


specific heat of water (s) = 1cal / g-c
heat energy transferred Q = ms (T2-T1)
Q = 1x1x(100-0)
= 100 cal


100 cal is the temperature





hope it helped you :)

plzz mark as best and also click on thank

Answer 2

The amount of heat transferred in the process is 100 cal.

Explanation:

The energy transferred during the process is given as

 $Q=mC_P(\Delta T)$

Here, m is the mass of the water, $C_P$ is the specific heat of water and  $\Delta T$ is the change in temperature.

Given m = 1 g,  $\Delta T=100 \ ^OC$ and we know, $C_P=1cal/g\ ^0C$ ( for water).

Substitute the given values, we get

$Q=1g\times1cal/g\ ^0C\times100\ ^0C$

Q = 100 cal.

Thus, the amount of heat transferred in the process is 100 cal.

#Learn More: Heat.

https://brainly.in/question/1298111