Question

how many grams of water were present if 6.5 KJ of heat energy raise the temperature from 34-56°C specific heat of water equals 4200 J/Gtimes Celsius

Answer 1

Here the formula we will use =

$\Rightarrow$ Q=mc$(T_2-T_1)$

where Q= Heat energy raised

m=mass of water

c=specific heat

$T_1$ =initial temperature

$T_2$ = final temperature

Now according to question

Q=6.5KJ=6500J

$T_1$ =34 degree celsius

$T_2$ =56 degree celsius

c=4200J/Gtimes Celsius

So from the above mention formulla

mass of water (m)= $\dfrac{Q}{c\times(T_2-T_1)}$

m=$\dfrac{6500}{4200\times(56-34)}$

m=0.07 gm of water.