Question
Exercises For the exercises in this chapter, use the value of \( \mathrm{c}=1.0 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C} \) for the specific heat of water. The conversion between calories and joules is \( 4.19 \mathrm{~J}=1 \mathrm{cal} \). E1. An object has a temperature of \( 22^{\circ} \mathrm{C} \). a. What is its temperature in degrees Fahrenheit? b. What is this temperature on the absolute (Kelvin) scale? E2. The temperature on a winter day is \( 20^{\circ} \mathrm{F} \). What is the tem- perature in degrees Celsius?
Ask by Barnett Mills. in the United States
Jan 13,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
**E1. An object at \( 22^{\circ} \mathrm{C} \):**
- **a.** \( 71.6^{\circ} \mathrm{F} \)
- **b.** \( 295.15 \, \mathrm{K} \)
**E2. A winter day at \( 20^{\circ} \mathrm{F} \):**
- \( \approx -6.67^{\circ} \mathrm{C} \)
Solution
Sure, let's tackle each exercise step by step.
---
### **E1. An object has a temperature of \( 22^{\circ} \mathrm{C} \).**
#### **a. What is its temperature in degrees Fahrenheit?**
To convert Celsius to Fahrenheit, use the formula:
\[
^\circ F = \left(^\circ C \times \frac{9}{5}\right) + 32
\]
**Given:**
\[
^\circ C = 22^{\circ} \mathrm{C}
\]
**Calculation:**
\[
^\circ F = (22 \times \frac{9}{5}) + 32
\]
\[
^\circ F = (22 \times 1.8) + 32
\]
\[
^\circ F = 39.6 + 32
\]
\[
^\circ F = 71.6^{\circ} \mathrm{F}
\]
**Answer:**
\[
71.6^{\circ} \mathrm{F}
\]
---
#### **b. What is this temperature on the absolute (Kelvin) scale?**
To convert Celsius to Kelvin, use the formula:
\[
K = ^\circ C + 273.15
\]
**Given:**
\[
^\circ C = 22^{\circ} \mathrm{C}
\]
**Calculation:**
\[
K = 22 + 273.15
\]
\[
K = 295.15 \, \mathrm{K}
\]
**Answer:**
\[
295.15 \, \mathrm{K}
\]
---
### **E2. The temperature on a winter day is \( 20^{\circ} \mathrm{F} \). What is the temperature in degrees Celsius?**
To convert Fahrenheit to Celsius, use the formula:
\[
^\circ C = \left(^\circ F - 32\right) \times \frac{5}{9}
\]
**Given:**
\[
^\circ F = 20^{\circ} \mathrm{F}
\]
**Calculation:**
\[
^\circ C = (20 - 32) \times \frac{5}{9}
\]
\[
^\circ C = (-12) \times \frac{5}{9}
\]
\[
^\circ C = -\frac{60}{9}
\]
\[
^\circ C \approx -6.67^{\circ} \mathrm{C}
\]
**Answer:**
\[
\approx -6.67^{\circ} \mathrm{C}
\]
---
Feel free to reach out if you have any more questions or need further clarification!
Answered by UpStudy AI and reviewed by a Professional Tutor
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Simplify this solution Mind Expander
To convert from degrees Celsius to Fahrenheit, you can use the formula \( F = \frac{9}{5}C + 32 \). So for \( 22^{\circ}C \), the calculation would be \( F = \frac{9}{5}(22) + 32 = 71.6^{\circ}F \). For the absolute (Kelvin) scale, use \( K = C + 273.15 \), giving \( 22 + 273.15 = 295.15 \, K \). To convert from Fahrenheit to Celsius, the formula to use is \( C = \frac{5}{9}(F - 32) \). Plugging in \( 20^{\circ}F \) gives \( C = \frac{5}{9}(20 - 32) = -6.67^{\circ}C \). Voilà, you’ve just calculated those temperatures!
