Example 16-12 depicts the following scenario. A 0.50 kg block of metal with an i
ID: 1568102 • Letter: E
Question
Example 16-12 depicts the following scenario. A 0.50 kg block of metal with an initial temperature of 50.7 C is dropped into a container holding 1.1 kg of water at 20.1 C. The specific heat of the metal block is 390 J/(kgK).
Problem 16.95-16.96 Example 16-12 depicts the following scenario. A 0.50 kg block of metal with an initial temperature of 50.7 Cis dropped into a container holding 1.1 kg of water at 20 The specific heat of the metal block is 390 J/(kg.K). Initial conditions Equilibrium Now we will consider some slightly different scenarios related to Example 16-12. Part A Refer to Example 16-12. Suppose the mass ofthe block is to be increased enough to make the final temperature of the system equal to 23.3 C. What is the required mass? Everything else in Example 16-12 remains the same. Express your answer using two significant figures. kg Submit My Answers Give U Part B Refer back to Example 16-12. Suppose the initial temperature of the block is to be increased enough to make the final temperature of the system equal to 22.5 C. What is the required initial temperature? Everything else remains the same as in Example 16-12. Express your answer using two significant figures. Submit My Answers Give UExplanation / Answer
heat lost by mass = heat gained by water
Mb*Sb*(Tb-T) = Mw*Sw*(T-Tw)
Mb*390*(50.7-23.3) = 1.1*4186*(23.3-20.1)
mass of the block = 1.38 kg
mass of the block is increased by 1.38 - 0.5 = 0.88 kg
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part B
heat lost by mass = heat gained by water
Mb*Sb*(Tb-T) = Mw*Sw*(T-Tw)
0.5*390*(Tb-22.5) = 1.1*4186*(22.5-20.1)
temperature of block Tb = 79.17 oC
temperature of the block is increased by 79.17 - 50.7= 24.87 oC
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