IDS 102 Quiz 1 Jiminy Cricket and mixing water. What do we know about  J? Celsius Temperature Jiminy Temperature Cool night 15  C 76  J Hot day 35.

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Presentation transcript:

IDS 102 Quiz 1 Jiminy Cricket and mixing water

What do we know about  J? Celsius Temperature Jiminy Temperature Cool night 15  C 76  J Hot day 35  C 220  J

1a) The Graph: (1 pt) This is good… This is good…

1a) The Graph: Is this more useful? (Maybe, maybe not) Is this more useful? (Maybe, maybe not)

1b) Find the equation (2 pts) First find the slope First find the slope Key question: What’s y? What’s x? Key question: What’s y? What’s x?

1b) The equation Then find the intercept Then find the intercept Choose ONE point on your graph… Choose ONE point on your graph… …such as (15  C, 76  J) (what’s y? what’s x?) …such as (15  C, 76  J) (what’s y? what’s x?) Notice that the intercept is NEGATIVE! We knew this from the graph.

1b) The equation So the equation is… So the equation is… (  Jiminy) = (7.2)(  Celsius) – 32 Hint: write it in a way so that you can’t forget: “what’s y? what’s x”?

1c) Specific temperatures (2 pts) In degrees Jiminy, water freezes at… In degrees Jiminy, water freezes at… (7.2)(0  Celsius) – 32 = -32 degrees Jiminy In degrees Jiminy, water boils at… In degrees Jiminy, water boils at… (7.2)(100  Celsius) – 32 = 688 degrees Jiminy

#2) Complete the sentence (1 pt) When we say that one object is warmer than another, what we mean is that the warmer object has a higher temperature. When we say that one object is warmer than another, what we mean is that the warmer object has a higher temperature. When a warm object is placed in contact with a cold object the warm object gives some heat to the cold object. When a warm object is placed in contact with a cold object the warm object gives some heat to the cold object.

#3) Mixing water a) (2 pts) If only 20 calories changes hands… a) (2 pts) If only 20 calories changes hands… The 50 gram water sample (initially at 20  ) warms up, so we just need to figure out how much: The 50 gram water sample (initially at 20  ) warms up, so we just need to figure out how much: The 20 gram water sample (initially at 55  C) cools down by The 20 gram water sample (initially at 55  C) cools down by

#3) Mixing water b) (2 pts) Once heat transfer has stopped… b) (2 pts) Once heat transfer has stopped… The final temperature will be The final temperature will be You could also solve this by preparing a table. The table would have to be large enough that a total of 500 calories could be transferred from the warm water to the cool water.