A 100 g aluminum calorimeter contains 250 g of water. The two substances are in

Question

A 100 g aluminum calorimeter contains 250 g of water. The two substances are in thermal equilibrium at 10degreeC. Two metallic blocks are placed in the water. One is a 50 g piece of copper at 80degreeC. The other sample has a mass of 75 g and is originally at a temperature of 100degreeC. The entire system stabilizes at a final temperature of 20degreeC. Determine the specific heat of the unknown second sample. A 40-g block of ice is cooled to -68degreeC and is then added to 590 g of water in an 80-g copper calorimeter at a temperature of 23degreeC. Determine the final temperature of the system consisting of the ice, water, and calorimeter. (If not all the ice melts, determine how much ice is left.) Remember that the ice must first warm to 0degreeC, melt, and then continue warming as water, (The specific heat of ice is 0.500 cal/g middot degreeC = 2,090 J/kg middot degreeC.) T_f = degreeC m_ice final = g

Explanation / Answer

Specific heat of aluminum = 0.897 J / g degC.
Heat gained by calorimeter = 100 * 10 * 0.897 J = 897 J

Specific heat of water = 4.18 J / g degC.
Heat gained by water = 250 * 10 * 4.18 J = 10450 J

Specific heat of copper = 0.385 J / g degC.
Heat loss by copper block = 50 * (66-10) * 0.385 J = 1078 J

Let specific heat of unknown block be x J / g degC.
Heat loss by unknown block = 75 * (100-10) * x J = 6750x J

Heat gain = Heat loss
897 + 10450 = 1078 + 6750x
6750x = 10269
x =~ 1.5213

Specific heat of unknown substance is 1.5213 J / g degC.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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