Answer :
The 100 grams of ice at 0 °C is placed in 500 grams of water at 50 °C. the final temperature is 29 °C .
The specific heat equation is given as :
q = mc ΔT
specific heat of water , c = 4.18 J/g °C
q = n ΔHfusion
heat of fusion of water = 6.01 kJ /mol
moles of water in 100 g of water = 100 / 18 = 5.55 mol
heat needed from solid 0 °C to liquid 0 °C .
q1 = 5.55 × 6.01 = 33.36 kJ
therefore, q1 + q2 = - q3
33.36 + mc ΔT = - mc ΔT
33.36 + 100 × 4.18 (T - 0) = - 500 × 4.18 (T - 25)
33.36 + 418 T = - 2090 ( T - 25 )
33.36 + 418 T = -2090 T + 52250
2508 T = 52216.6
T = 29 °C
Thus, The 100 grams of ice at 0 °C is placed in 500 grams of water at 50 °C. the final temperature is 29 °C
To learn more about specific heat here
https://brainly.com/question/11297584
#SPJ4
