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Chapter 3: Q46E (page 576)

An open container holds 0.550kg of ice at -15.0掳C. The mass of the container can be ignored. Heat is supplied to the container at the constant rate 800.0 J/min of for 500.0 min. (a) After how many minutes does the ice start to melt? (b) After how many minutes, from the time when the heating is first started, does the temperature begin to rise above 0.0掳C? (c) Plot a curve showing the temperature as a function of the elapsed time.

Short Answer

Expert verified

(a) t = 21.7 min

(b) t = 251.7 min

(c)

Step by step solution

01

Calculate the time using the heat flow equation

The heat flow equation is Q=mc鈭�T.

Given that mice=0.550kg,Tice=-15掳Cand鈭�Q鈭�t=800J/min.

Q=尘肠未罢Q=0.550脳2110脳(0+15)Q=1.73脳104J

By solving the rate localid="1665117914592" 鈭�Q鈭�tfor time t

t=Q螖Q螖tt=1.73脳104800t=21.7min

So, the ice started to melt after 21.7 min.

02

Calculate the time for vaporisation

The heat involved in the vaporization process is

Q=mLfQ=0.55脳3.34脳105Q=1.84脳105J

By solving the time for the rate

t2=Q螖Q螖tt2=1.84脳105800t2=230min

So, by summing the minutes at which ice starts melting and the time at which solid converts to liquid, we will get the total time

t1+t2=21.7+230t1+t2=251.7min

03

Plot the graph

The graph of the process will be

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Most popular questions from this chapter

Imagine a special air filter placed in a window of a house. The tiny holes in the filter allow only air molecules moving faster than a certain speed to exit the house, and allow only air molecules moving slower than that speed to enter the house from outside. What effect would this filter have on the temperature inside the house? (It turns out that the second law of thermodynamics鈥攚hich we will discuss in Chapter 20鈥攖ells us that such a wonderful air filter would be impossible to make.)

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