Question

The same amount of heat transferred into identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of steel, originally at 20.00; c = 0.108 kcal/kg 0C .

29.3 0 C


26.9 0 C


32.3 0 C


27.4 0 C

Answer 1

Apply the formula

`\boxed{\sf Q=mc\Delta T}`

`\\ \tt{:}\longrightarrow \Delta T=(Q)/(mc)`

`\\ \tt{:}\longrightarrow \Delta T=(1)/(0.108)`

`\\ \tt{:}\longrightarrow T_f-20=9.3`

`\\ \tt{:}\longrightarrow T_f=29.3°C`

Answer 2

The assumption is Originally at 20°C

Now

• m=1
• Q=1
• c=0.108

We need ∆T

• Q=mc∆T
• 1=1(0.108)∆T
• 0.108∆T=1
• ∆T=1/0.108=9.26°C

Now

• ∆T=T_f-20
• 9.26=T_f-20
• T_f=29.26°C

Option A