If 500 grams of water at 25°C loses 10500 Joules of heat,?

Question

what wil be the final temperature of the water? c water = 4.18 J / g °C:
and how are you coming to that conclusion
my answer does not look right to me

Answer 1

This is a classical Q=mc(DeltaT) question. If you replace DeltaT with T2-T1 and then solve for T2, you should get your answer.

Q=mc(DeltaT)
=mc(T2-T1)
Q/(mc)=T2-T1
Q/)mc) + T1=T2

Q=-10500J
m=500g
c=4.18 J / (g °C)
T1=25°C

You get a final answer of 19.98 °C

Answer 2

20 degrees C

Answer 3

1g of water = 4.18J/g/°C

500 x 4.18 = 2,090J/°C
10,500J ÷ 2090 = 5.024 °C. (5°C to nearest °)
Final temp. = 25 - 5 = 20°C