Question

** worth 20 points + brainliest **

Answer 1

Answer : The correct option is, (D) $83^oC$

Solution :

Formula used :

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

where,

Q = heat released = -1300 J

m = mass of water = 40 g

c = specific heat of water = $4.18J/g^oC$      

$\Delta T=\text{Change in temperature}$  

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $91.0^oC$

Now put all the given values in the above formula, we get the final temperature of water.

$-1300J=40g\times 4.18J/g^oC\times (T_{final}-91.0^oC)$

$T_{final}=83.2^oC\approx 83^oC$

Therefore, the final temperature of the water is, $83^oC$