Question

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

Answer 1

Explanation:

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