Question

a) how much heat must be absorbed by ice of mass $m_{i}=720g$ at $t_{i}=-10.0^{circ}c$ to take it to the liquid state at $t_{f}=15.0^{circ}c$? ( $c_{i}=2090j/(kgcdot k), c_{w}=4187j/(kgcdot k), c_{v}=1520j/(kgcdot k), l_{f}=2220j/kg, l_{v}=334\times10^{3}j/kg$ )
a) $q = q_{1}+q_{2}+q_{3}$
ice $\to - 10.0^{circ}c$
$downarrow q_{1}=m_{i}c_{i}(0 - (-10.0))$
ice $\to 0^{circ}c$
$downarrow q_{2}=m_{i}l_{f}$
water $\to 0^{circ}c$
$downarrow q_{3}=m_{i}c_{w}(15.0 - 0)$
water $\to 15.0^{circ}c$
b) find the total energy needed to vaporize all mass of the ice?
water $\to 15.0^{circ}c$
$downarrow q_{4}=m_{i}c_{w}(85.0)$
water $\to 100^{circ}c$
$downarrow q_{5}=m_{i}l_{v}$
vapor $\to 100^{circ}c$
$q = m_{i}c_{i}\times100 + m_{i}l_{f}+m_{i}c_{w}\times15.0 + m_{i}c_{w}\times85.0 + m_{i}l_{v}$
part(a) $+(0.720\times4187\times85.0)+(0.720\times334\times10^{3})$
$q =? j$

Explanation:

Step1: Calculate heat to raise ice temperature to 0°C

$Q_1 = m\times c_i\times\Delta T$, where $m = 0.72\ kg$ (since $720\ g=0.72\ kg$), $c_i = 2090\ J/(kg\cdot K)$ and $\Delta T=(0 - (- 10))\ K = 10\ K$. So $Q_1=0.72\times2090\times10 = 15048\ J$.

Step2: Calculate heat for phase - change from ice to water

$Q_2=m\times L_f$, with $m = 0.72\ kg$ and $L_f = 334\times10^{3}\ J/kg$. So $Q_2=0.72\times334\times10^{3}=240480\ J$.

Step3: Calculate heat to raise water temperature from 0°C to 15°C

$Q_3 = m\times c_w\times\Delta T$, where $m = 0.72\ kg$, $c_w = 4187\ J/(kg\cdot K)$ and $\Delta T=(15 - 0)\ K = 15\ K$. So $Q_3=0.72\times4187\times15=45219.6\ J$.

Step4: Calculate total heat for part (a)

$Q_{total}=Q_1 + Q_2+Q_3=15048+240480 + 45219.6=300747.6\ J$.

Step5: Calculate heat to raise water temperature from 15°C to 100°C for part (b)

$Q_4=m\times c_w\times\Delta T$, with $m = 0.72\ kg$, $c_w = 4187\ J/(kg\cdot K)$ and $\Delta T=(100 - 15)\ K = 85\ K$. So $Q_4=0.72\times4187\times85 = 257379\ J$.

Step6: Calculate heat for phase - change from water to vapor

$Q_5=m\times L_v$, where $m = 0.72\ kg$ and $L_v = 2260\times10^{3}\ J/kg$. So $Q_5=0.72\times2260\times10^{3}=1627200\ J$.

Step7: Calculate total heat for part (b)

$Q_{total - b}=Q_{total}+Q_4 + Q_5=300747.6+257379+1627200=2185326.6\ J$.

Answer:

a) $300747.6\ J$
b) $2185326.6\ J$