# Find the total heat ….

Calculate the total heat (in J) needed to convert 9.01g (0.500 moles) of ice at -10.00 oC to liquid water at 5.00 oC. The melting point of water is 0 oC ; Δ H fusion for water is 6.02 kJ/mole ; C solid is 2.09 J/g oC ; C liquid is 4.21 J/g*oC.

The total heat needed to convert the ice into water is equal to the heat needed to rise the temperature from $-10$ C to $0$ C (Q1) plus the heat needed to melt the ice into water at $0$ C (Q2) plus the heat needed to rise the temperature of the water from $0$ C to $5$ C (Q3)

$Q =Q1 + Q2+ Q3$

$Q1 =m*c1*\Delta(T1),$

where m is the mass of ice, c1 is the caloric capacity of the ice and $\Delta(T1)$ the variation of the ice temperature

$Q2 = m*\Delta(H)$, where $\Delta(H)$ is the latent fusion heat of the ice (in J/g !!!! $=334$ J/g)

$Q3 = m*c2*\Delta(T2)$

where $m$ is the mass of the water, $c2$ is the caloric capacity of the water and $\Delta(T2)$ is the variation of the water temperature

$Q =m*c1*\Delta(T1) +m*\Delta(H) + m*c2*\Delta(T2) =$

$= 9.01*2.09*(0-(-10)) +9.01*334 + 9.01*4.21*(5-0) = 3387.31 J$