Answer :
Answer:
Lead shows the greatest temperature change upon absorbing 100.0 J of heat.
Explanation:
[tex]Q=mc\Delte T[/tex]
Q = Energy gained or lost by the substance
m = mass of the substance
c = specific heat of the substance
ΔT = change in temperature
1) 10.0 g of copper
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the copper = c = 0.385 J/g°C
[tex]\Delta T=\frac{Q}{mc}[/tex]
[tex]=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC[/tex]
2) 10.0 g of aluminium
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the aluminium= c = 0.903 J/g°C
[tex]\Delta T=\frac{Q}{mc}[/tex]
[tex]=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC[/tex]
3) 10.0 g of ethanol
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the ethanol= c = 2.42 J/g°C
[tex]\Delta T=\frac{Q}{mc}[/tex]
[tex]=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC[/tex]
4) 10.0 g of water
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the water = c = 4.18J/g°C
[tex]\Delta T=\frac{Q}{mc}[/tex]
[tex]=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC[/tex]
5) 10.0 g of lead
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the lead= c = 0.128 J/g°C
[tex]\Delta T=\frac{Q}{mc}[/tex]
[tex]=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC[/tex]
Lead shows the greatest temperature change upon absorbing 100.0 J of heat.