The melting point of aluminum is $660 ^ { \circ } \mathrm { C }$ its latent heat of fusion is $4.00 \times 10 ^ { 5 } \mathrm {~J} / \mathrm { kg }$ and its specific heat is $900 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ If 300 kJ of heat are added to 442 g of aluminum at $100 ^ { \circ } \mathrm { C }$
what is the final state of the system? That is, how much is liquid, how much is solid, and
what is its temperature?

Question

The melting point of aluminum is $660 ^ { \circ } \mathrm { C }$ its latent heat of fusion is $4.00 \times 10 ^ { 5 } \mathrm {~J} / \mathrm { kg }$ and its  specific heat is $900 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ If 300 kJ of heat are added to 442 g of aluminum at $100 ^ { \circ } \mathrm { C }$
what is the final state of the system? That is, how much is liquid, how much is solid, and
what is its temperature?

Quizplus · Accepted Answer

The Answer of The melting point of aluminum is $660 ^ { \circ } \mathrm { C }$ its latent heat of fusion is $4.00 \times 10 ^ { 5 } \mathrm {~J} / \mathrm { kg }$ and its  specific heat is $900 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ If 300 kJ of heat are added to 442 g of aluminum at $100 ^ { \circ } \mathrm { C }$
what is the final state of the system? That is, how much is liquid, how much is solid, and
what is its temperature?

The Melting Point of Aluminum Is 660∘C660 ^ { \circ } \mathrm { C }660∘C

The Melting Point of Aluminum Is $660 ^ { \circ } \mathrm { C }$