Question

10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C 10.0 g Pb C Pb = 0.128 J/g°C Which of the following substances (with specific heat capacity provided) would show the greatest temperature change upon absorbing 100.0 J of heat

Answer 1

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Step-by-step explanation:

`Q=mc\Delte T`

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

`\Delta T=(Q)/(mc)`

`=(100.0 J)/(10 g* 0.385J/g^oC)=25.97^oC`

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

`\Delta T=(Q)/(mc)`

`=(100.0 J)/(10 g* 0.903 J/g^oC)=11.07^oC`

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

`\Delta T=(Q)/(mc)`

`=(100.0 J)/(10 g* 2.42 J/g^oC)=4.13 ^oC`

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

`\Delta T=(Q)/(mc)`

`=(100.0 J)/(10 g* 4.18 J/g^oC)=2.39 ^oC`

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

`\Delta T=(Q)/(mc)`

`=(100.0 J)/(10 g* 0.128 J/g^oC)=78.125^oC`

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.