Use the information below to answer the following problems. A 500 g iron kettle

Question

Use the information below to answer the following problems. A 500 g iron kettle is tilled with 270 g of liquid water, both at 15 degree C. When placed on si hot stove, the water will eventually boil. How much total heat (in kJ) is required to bring the iron kettle and the water to 100 degree C ? What percentage of this total heat is required to heat only the iron kettle? If a copper kettle (same number of moles) was used instead of the iron kettle, would MORE. LESS, or the SAME amount of total heat be required to bring the kettle to a boil?

Explanation / Answer

q water = m * s * dT + dH fusion * number of moles
q water = 270 * 4.18 * (100 - 15) + 40.7 * 270/18
q water = 96.54 kJ
q iron = m * s * dT
q iron = 500/55.8 * 0.029*10^3 * (100 - 15)
q iron = 220.87 kJ
Q total = 96.54 + 220.87 = 317.41 kJ
percentage of q iron = 220.87/317.41*100 = 69.58%
for copper
Q = 500 * 0.385 *(100-15)
Q = 16.362 kJ

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