Question

An iron bar that loses 2.92kj when it’s temperature decreases from 254c to 72c

Answer 1

`Q= m * c * \Delta T`

where

`\Delta T` = Final T - Initial T

Q is the heat energy in calories = 2.92 KJ = 2920 Joules

c is the specific heat capacity (for iron 0.450 J/(g℃))

m is the mass of water

Plugging in the values

`2920J=m*0.450 \frac {J}{(g^\circ{C})}*(254^\circ{C}-72^\circ{C})`

`2920J=m*0.450 \frac {J}{(g^\circ{C})}*(182^\circ{C})`

`mass=\frac{2920 J}{0.450 \frac{J}{g^\circ{C}} * 182^\circ{C}}`

mass = 35.6℃ is the Answer