![]() ![]() We assume no loss of heat energy to the outside during the transfer. We assume all heat absorbed by the water was lost by the metal. ![]() I used 50.0 g because the density of water is 1.00 g/mL and I had 50.0 mL of water. (c) What is the specific heat capacity of the metal? (b) How many joules of energy did the metal lose? (a) How many joules of energy did the water absorb? The temperature of the water rose to 28.1☌. Problem #6: A 12.48 g sample of an unknown metal is heated to 99.0 ☌ and then was plunged into 50.0 mL of 25.0 ☌ water. That means the ice-water mix remained at zero Celsius as the 109.5 g of ice melted.ģ) Determine aomic weight of the element: Note: we know the change in temperature is 153.0 ☌ because there is still ice in the water. What is the specific heat of the metal in J/g-☌? Given the molar heat capacity of the metal is 26.31 J/mol ☌, what is the atomic weight and identity of the metal?ġ) Determine energy needed to melt the ice: 109.5 g of ice melts and there is still an ice-water mixture. Problem #5: A 500.0 g sample of an element at 153.0 ☌ is dropped into an ice-water mixture. ![]() We know this from the presence of the ice.ģ) Determine the specific heat of the metal: Also, notice that the water was at zero ☌. Note the inclusion of the melted 15 g of ice. Note that the 100 g of water is not mentioned yet.Ģ) Determine heat need to raise 115 g of water from 0 to 10.0 ☌: What is the specific heat of the metal?ġ) Determine heat required to melt the ice: All the ice melted and the temperature in the container rose to 10.0 ☌. ![]() Problem #4: A 35.0 g block of metal at 80.0 ☌ is added to a mixture of 100.0 g of water and 15.0 g of ice in an isolated container. Verrrry interesting!ġ) Determine heat gained by the ice that melted:Ģ) Substitue and solve for the specific heat: Rather, some ice melts and the whole ice-water system stays at zero Celsius. What is the specific heat of the metal? (The heat of fusion of ice = 334.166 J g¯ 1.)Ĭomment: this variation of the usual suspects (detailed above) does NOT involve a temperature change in the water, only in the metal. After the system had reached equilibrium it was determined that 9.15 g of the ice had melted. Problem #3: A 43.2 g block of an unknown metal at 89.0 ☌ was dropped into an insulated vesssel containing 43.00 g of ice and 26.00 g of water at 0 ☌. Notice that parts (1) and (2) are the equivalent of q lost = q gained and that (3) is the usual answer sought in problems of this type. That is because the question is broken up into four parts. How many joules of energy did the metal lose? How many joules of energy did the water absorb?Ģ. Assuming no loss of energy to the surroundings:ġ. The temperature of the water rose to 28.1 ☌. Problem #2: A 12.48 g sample of an unknown metal, heated to 99.0 ☌ was then plunged into 50.0 mL of 25.0 ☌ water. (21.5) (100 − x) (0.449) = (132.0) (x − 20) (4.184)Ī) 100 − x is the Δt for the metal it starts at 100.0 ☌ and drops to some unknown, final value.ī) x − 20 is the Δt for the water it starts at 20.0 ☌ and rises to some unknown, final value.Ĭ) Since both metal and water wind up at the same ending value, we need to use only one unknown for the two Δt expressions. What will be the final temp of the system? Specific heat of iron is 0.449 kJ/kg K. The mass of the water is 132.0 g and its temperature before adding the iron is 20.0 ☌. Problem #1: Suppose a piece of iron with a mass of 21.5 g at a temp of 100.0 ☌ is dropped into an insulated container of water. How to Determine the Specific Heat of a Substance ChemTeam: How to Determine Specific Heat: Problem 1 - 10 ![]()
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