Problem : A 500 g sample of an unknown material requires 750 calories of heat to

Question

Problem :
A 500 g sample of an unknown material requires 750 calories of heat to raise its temperature by 50 degrees C. Identify the material
Q=mc(triangle symbol)T
Question:
Have to use table below to answer question but not sure what it is asking I think the 750 calories is Q and 500g is m but not sure what to do after that


specific heats of various subtances (at 20degreesC and 1 atm)
Substance Specific heat (cal/g)

Air (50C) 0.25
Alcohol, ethyl 0.58
Aluminum 0.22
Copper 0.093
Glass 0.20
Ice (-5C) 0.50
Iron or steel 0.11
Lead 0.031
Soil (average) 0.25
Steam (110C) 0.48
Water(15C) 1.00
Wood (average) 0.40

Explanation / Answer

You are correct. Trust your instincts and plug given values into equation.

q=mct where t=t final-t initial or difference in temperature.

q=750 cal, m=500 g, c is unknown, and t = 50 C

750=500c(50)

750=25000c

c=0.03

based on the table posted, lead's capacity is 0.031 and is the closest value to the derived answer.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.