Temperature and Heat Conversions Practice Answer Key (CHEM 1405)

Back to Chemistry 1405 Practice Problems

 

Convert the following to Fahrenheit:

1)      10.o C   ________          

LaTeX: ^{\circ}F=\frac{9}{5}^{\circ}C+32=\frac{9}{5}(10^{\circ}C)+32=50^{\circ}C∘F=95∘C+32=95(10∘C)+32=50∘C

 

2)      30.o C   ________

LaTeX: ^{\circ}F=\frac{9}{5}^{\circ}C+32=\frac{9}{5}(30^{\circ}C)+32=86^{\circ}F∘F=95∘C+32=95(30∘C)+32=86∘F

 

3)      0o C     ________

LaTeX: ^{\circ}F=\frac{9}{5}^{\circ}C+32=\frac{9}{5}(0^{\circ}C)+32=32^{\circ}F∘F=95∘C+32=95(0∘C)+32=32∘F

 

Convert the following to Celsius:

4)      32o F   ________

LaTeX: ^{\circ}C=\frac{5}{9}(^{\circ}F-32)=\frac{5}{9}(32^{\circ}F-32)=0^{\circ}C∘C=59(∘F−32)=59(32∘F−32)=0∘C

 

5)      90.o F   ________

LaTeX: ^{\circ}C=\frac{5}{9}(^{\circ}F-32)=\frac{5}{9}(90^{\circ}F-32)=32^{\circ}C∘C=59(∘F−32)=59(90∘F−32)=32∘C

 

6)      212o F ________

LaTeX: ^{\circ}C=\frac{5}{9}(^{\circ}F-32)=\frac{5}{9}(212^{\circ}F-32)=100.^{\circ}C∘C=59(∘F−32)=59(212∘F−32)=100.∘C

 

Convert the following to Kelvin:

7)      0o C     ________

LaTeX: K=^{\circ}C+273=0^{\circ}C+273=273KK=∘C+273=0∘C+273=273K

 

8)      -50.o C ________

LaTeX: K=^{\circ}C+273=-50^{\circ}C+273=223KK=∘C+273=−50∘C+273=223K

 

9)      90.o C   ________

LaTeX: K=^{\circ}C+273=90^{\circ}C+273=363KK=∘C+273=90∘C+273=363K

 

10)   -20.o C  ________

LaTeX: K=^{\circ}C+273=-20^{\circ}C+273=253KK=∘C+273=−20∘C+273=253K

Convert the following to Celsius:

11)   112 K   ________

LaTeX: ^{\circ}C=K-273=112K-273=-161^{\circ}C∘C=K−273=112K−273=−161∘C

 

12)   200. K   ________

LaTeX: ^{\circ}C=K-273=200.K-273=-73^{\circ}C∘C=K−273=200.K−273=−73∘C

 

13)   273 K   ________

LaTeX: ^{\circ}C=K-273=273K-273=0^{\circ}C∘C=K−273=273K−273=0∘C

 

14)   350. K   ________

LaTeX: ^{\circ}C=K-273=350.K-273=77^{\circ}C∘C=K−273=350.K−273=77∘C

 

 

Use the information below to complete the following problems.

 LaTeX: 1cal=4.184J1cal=4.184J                        LaTeX: 1000J=1kJ1000J=1kJ

 

15)  Calculate the number of joules in 124 calories.

LaTeX: 124cal\times(\frac{4.184J}{1cal})=519J124cal×(4.184J1cal)=519J

 

16)  Calculate the number of calories in 1.43 kJ.

LaTeX: 1.43kJ\times(\frac{1000J}{1kJ})\times(\frac{1cal}{4.184J})=342cal1.43kJ×(1000J1kJ)×(1cal4.184J)=342cal

 

 

17)  Calculate the number of kJ in 784,500 J.

LaTeX: 784,500J\times(\frac{1kJ}{1000J})=784.5kJ784,500J×(1kJ1000J)=784.5kJ

 

 

18)  Calculate the number of calories in 546 J.

LaTeX: 546J\times(\frac{1cal}{4.184J})=130. cal546J×(1cal4.184J)=130.cal