Question

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

Answer 1

Answer:

$Q= m \times c \times \Delta T$

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  

$2920J=m\times0.450 \frac {J}{(g^\circ{C})}\times(254^\circ{C}-72^\circ{C})$

$2920J=m\times0.450 \frac {J}{(g^\circ{C})}\times(182^\circ{C})$

$mass=\frac{2920 J}{0.450 \frac{J}{g^\circ{C}} \times 182^\circ{C}}$

mass = 35.6℃ is the Answer