150.0 grams of iron at 95.0 °C, is placed in an insulated container containing 500.0 grams of water. The temperature of the water increases to 27.2°C. What was the initial temperature of the water? The specific heat of water is 4.18 J/g °C and the specific heat of iron is 0.444 J/g °C

Question

Frances Garrison

Chemistry

21 May, 22:20

150.0 grams of iron at 95.0 °C, is placed in an insulated container containing 500.0 grams of water. The temperature of the water increases to 27.2°C. What was the initial temperature of the water? The specific heat of water is 4.18 J/g °C and the specific heat of iron is 0.444 J/g °C

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Fonzie · Answer 1

in it I mass=150g, initial temp 1=95, mass2=500g, temp2=?, final temperature=27.2 C1=0.444C2=4.18

using formula

m1c1 (final temp-initial temp1) = m2c2 (temp2-final temp)

150x0.444 (27.2-95) = 500x4.18 (?-27.2)

66 (-67.8) = 2000 (?-27.5)

-4474.8=2000?-55000

collect like terms

-4474.8+55000=2000?

50525=2000?

divide both sides by 2000

2000?/2000=50525/2000

initial temperature = 25.26 degree Celsius