Answer:
Explanation:
we know that specific heat is the amount of heat required to raise the temperature of substance by one degree mathmeticaly
Q=mcΔT
ΔT=T2-T1
ΔT=26.8-10.2=16.6
C for water is 4.184
therefore
Q=1.00*4.184*16.6
Q=69.4 j
now we have to covert joule into calorie
1 calorie =4.2 j
x calorie=69.4 j/2
so 69.4 j =34.7 calorie thats why 34.7 calorie heat is required to raise the temperature of water from 10.2 to 26.8 degree celsius
