Exploring Creation with Chemistry 2nd Edition Practice Problems #2

The temperature of deep space is 3.5 K. What is that in Celsius? In Fahrenheit?

-269.5 C = -453.1 F

How many Joules does it take to heat up 15.1 kg of glass from 15 C to 45 C? (refer to Table 2.1)

15.1 kg = 1.51 x 10⁴ g q = (15.1 x 10⁴ g)·(0.8372 J/g.C)·(45 C - 15 C) q = (15.1 x 10⁴ g)·(0.8372 J/g.C)·(3.0 x 10¹ C) = 3.8 x 10⁵ J

The average person uses 2,500.0 Cal of energy per day. How many Joules is that? Remember that a food calorie (Cal) is 1,000 chemistry calories (cal).

2,500.0 C = 2.5000 x 10⁶ c · 4.184 J = 1.046 x 10⁷ J

Some parts of the western United States regularly reach temperatures of 115 F in the summer. What is that in C?

5/9·(115-32)=46.1 C

Convert 15.0 C to Fahrenheit

9/5(15.0)+32 = 59.0 F

124.1 g of an unknown substance absorbs 50.0 kJ of heat and increases its temperature by 26.3 C. What is its specific heat? (Remember that "k" is the abbreviation for "kilo," so "kJ" stands for kiloJoules.)

c = 50.0 kJ/(124.1 g)·(36.3 C) = 0.0111 kJ/g. C

A calorimeter is filled with 150.0g of water at 24.1 C. A 50.0g sample of a metal at 100.0 C is dropped in this calorimeter and causes the temperature to increase a total of 5.4 C. What is the specific heat of the metal? Ignore the calorimeter in this problem.

q = (150.o g)·(4.184 J / g. C)·(5.4) q = 3.4 x 10³ J -qᵒᵇʲᵉᶜᵗ = qʷᵃᵗᵉʳ + qᶜᵃᶫᵒʳᶦᵐᵉᵗᵉʳ -qᵒᵇʲᵉᶜᵗ = 3.4 x 10³ J c = -3.4 x 10³ J/(50.0 g)·(29.5 C - 100.0 C) = 0.96 J/g. C

Review question #7 mentioned that drinking ice-cold water is a way of burning excess Calories. Calculate how many Calories are burned when a 12-ounce (3.4 x 10² g) glass of water at 0.0 C is warmed up to body temperature (37.0 C).

q = (3.4 x 10² g)·(1.000 cal/g. C)·(37.0 C) q = 1.26 x 10⁴ cal q = 12.6 Cal

A 345.1 g sample of copper at 100.0 C is dropped into a 4.5 g calorimeter made of an unknown substance. If the calorimeter has 150.0 grams of water in it and the temperature changed from 24.2 C to 25.1 C, what is the specific heat of the calorimeter? (refer to Table 2.1)

q = (345.1 g)·(0.3851 J/g. C)·(25.1 C - 100.0 C) q = -9950 J -qᵒᵇʲᵉᶜᵗ = qʷᵃᵗᵉʳ + qᶜᵃᶫᵒʳᶦᵐᵉᵗᵉʳ -(-9950 J) = 600 J + qᶜᵃᶫᵒʳᶦᵐᵉᵗᵉʳ qᶜᵃᶫᵒʳᶦᵐᵉᵗᵉʳ = 9400 J 9400 J/(4.5 g)·(0.9 C) = 2000 J/g. C

A 245 g piece of copper at room temperature (25 C) loses 456.7 Joules of heat. What will its new temperature be?

ΔT = -456.7 J/(245 g )·(0.3851 J/g. C) = -4.84