Respuesta :
Answer:627j
Explanation:
H=mcΔT
ΔH= 10*4.18*(300-285)
ΔH=10*4.18*15
ΔH=627j
The heat of the given water sample has been 0.627 kJ.
The specific heat has been described as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius. The expression for specific heat has been given as:
Heat = mass × specific heat × change in temperature
For the given water sample:
Specific heat = 4.18 J/K/g
Mass = 10 g
Initial temperature ([tex]T_i[/tex]) = 285 K
Final temperature ([tex]T_f[/tex]) = 300 K
Change in temperature ([tex]\rm \Delta T[/tex]) can be given as:
[tex]\Delta T =T_f\;-\;T_i\\\Delta T=300\;\text K\;-\;285\;\text K\\\Delta T=15\;\text K[/tex]
Substituting the values for heat:
[tex]\rm Heat=10\;g\;\times\;4.18\;J/K/g\;\times\;15\;K\\Heat\;=\;627.6\;J\\Specific\;heat\;=\;0.627\;kJ[/tex]
The heat of the given water sample has been 0.627 kJ.
For more information about the specific heat, refer to the link:
https://brainly.com/question/1209542
