Question

how many calories are required to heat 5.00 g of ice at -25.0 ^ { circ } c to 37.0 ^ { circ } c? use the conversions given below. first, the ice must be heated from -25.0 ^ { circ } c up to 0.0 ^ { circ } c, a change of 1 second, the ice at 0.0 ^ { circ } c must be melted to water at 2, there is no change in temperature third, the water must be heated from 0.0 ^ { circ } c to 37.0 ^ { circ } c, a change of 3 the answer is the addition of these three processes ( 0.504 mathrm { cal } / mathrm { g } ^ { circ } mathrm { c } ) ( 5.00 mathrm { g } ) ( 4 ) = 5 mathrm { cal } ( 79.9 mathrm { cal } / mathrm { g } ) ( 5.00 mathrm { g } ) = 6 mathrm { cal } ( 7 ) ( 5.00 mathrm { g } ) ( 8 ) = 9 mathrm { cal } 10 mathrm { cal } a. 0.504 mathrm { cal } / mathrm { g } ^ { circ } mathrm { c } b. 79.9 mathrm { cal } / mathrm { g } c. 0.999 mathrm { cal } / mathrm { g } ^ { circ } mathrm { c } d. 539 mathrm { cal } / mathrm { g } e. 0.497 mathrm { cal } / mathrm { g } ^ { circ } mathrm { c } f. 150.0 ^ { circ } mathrm { c } g. 100.0 ^ { circ } mathrm { c } h. 50.0 mathrm { g } i. 50.0 ^ { circ } mathrm { c } j. 96.0 ^ { circ } mathrm { c } k. 4.0 ^ { circ } mathrm { c } l. 25.0 ^ { circ } mathrm { c } m. 37.0 ^ { circ } mathrm { c } n. 5.00 mathrm { g } o. 647 p. 33.0 q. 1,243 r. 63.0 s. 4,795 t. 26,950 u. 399.5 v. 184.8 w. 32,988 x. 15.0 ^ { circ } mathrm { c } y. - 25.0 ^ { circ } mathrm { c } z. 5.00 aa. 0.0 ^ { circ } mathrm { c }

Explanation:

Step1: Calculate the temperature change for heating ice

The temperature change for heating ice from \(-25.0^{\circ}C\) to \(0.0^{\circ}C\) is \(\Delta T_1=0 - (- 25.0)=25.0^{\circ}C\). So, blank 1 is \(L\).

Step2: Melting of ice

When ice melts, it changes to water at \(0.0^{\circ}C\). So, blank 2 is \(AA\).

Step3: Calculate the temperature change for heating water

The temperature change for heating water from \(0.0^{\circ}C\) to \(37.0^{\circ}C\) is \(\Delta T_3 = 37.0-0.0 = 37.0^{\circ}C\). So, blank 3 is \(M\).

Step4: Calculate heat for heating ice

Using the formula \(Q = mc\Delta T\), where \(c = 0.504\space cal/g^{\circ}C\) (given for ice), \(m = 5.00\space g\), \(\Delta T=25.0^{\circ}C\). So, blank 4 is \(L\) ( \(25.0^{\circ}C\) ), and \(Q_1=(0.504)(5.00)(25.0)=63.0\space cal\). So, blank 5 is \(R\).

Step5: Calculate heat for melting ice

Using the formula \(Q=mL_f\), where \(L_f = 79.9\space cal/g\) (given for ice - water phase change), \(m = 5.00\space g\). So, \(Q_2=(79.9)(5.00)=399.5\space cal\). So, blank 6 is \(U\).

Step6: Calculate heat for heating water

Using the formula \(Q = mc\Delta T\), where \(c = 0.999\space cal/g^{\circ}C\) (specific - heat of water), \(m = 5.00\space g\), \(\Delta T = 37.0^{\circ}C\). So, blank 7 is \(C\), blank 8 is \(M\) ( \(37.0^{\circ}C\) ), and \(Q_3=(0.999)(5.00)(37.0)=184.8\space cal\). So, blank 9 is \(V\).

Step7: Calculate total heat

\(Q_{total}=Q_1 + Q_2+Q_3=63.0 + 399.5+184.8 = 647.3\approx647\space cal\). So, blank 10 is \(O\).

Answer:

1. \(L\)
2. \(AA\)
3. \(M\)
4. \(L\)
5. \(R\)
6. \(U\)
7. \(C\)
8. \(M\)
9. \(V\)
10. \(O\)