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Author Question: How much heat must be added to a 8.0-kg block of ice at -8C to change it to water at The specific ... (Read 53 times)

rosent76

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on: Oct 11, 2019

Question 1

A .20-kg ice cube at 0.0°C has sufficient heat added to it to cause total melting, and the resulting water is heated to How much heat is added? For water LF = 334,000 J/kg, LV = 2.256 × 106 J/kg, the c = 4.186 x 103 J/kg ∙ C.
◦ 81 kJ
◦ 14,000 kJ
◦ 59 kJ
◦ 130 kJ

Question 2

How much heat must be added to a 8.0-kg block of ice at -8°C to change it to water at  The specific heat of ice is 2050 J/kg ∙ °C, the specific heat of water is 4186 J/kg ∙ °C, the latent heat of fusion of ice is 334,000 J/kg, and 1 cal = 4.186 J.
◦ 140 kcal
◦ 810 kcal
◦ 730 kcal
◦ 780 kcal
◦ 180 kcal
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Marked as best answer by rosent76 on Oct 11, 2019

essyface1

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Reply #1 on: Oct 11, 2019
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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rosent76

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Reply #2 on: Oct 11, 2019
Thank you!


essyface1

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Reply #3 on: Oct 11, 2019
Always glad to help...


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