Answer:
Pb is the substance that experiments the greatest temperature change.
Explanation:
The specific heat capacity refers to the amount of heat energy required to raise in 1 degree the temperature of 1 gram of substance. The highest the heat capacity, the more energy it would be required. These variables are related through the equation:
Q = c . m . ΔT
where,
Q is the amount of heat energy provided (J)
c is the specific heat capacity (J/g.°C)
m is the mass of the substance
ΔT is the change in temperature
Since the question is about the change in temperature, we can rearrange the equation like this:
[tex] \Delta T = \frac{Q}{c.m}[/tex]
All the substances in the options have the same mass (m=10.0g) and absorb the same amount of heat (Q=100.0J), so the change in temperature depends only on the specific heat capacity. We can see in the last equation that they are inversely proportional; the lower c, the greater ΔT. Since we are looking for the greatest temperature change, It must be the one with the lowest c, namely, Pb with c = 0.128 J/g°C. This makes sense because Pb is a metal and therefore a good conductor of heat.
Its change in temperature is:
[tex] \Delta T = \frac{q}{c.m} = \frac{100.0 J}{0.128 J/g.C . 10.0g } = 78.1[/tex]
Pb would show the greatest temperature change upon absorbing 100.0 J of heat , which is : 78.125 ° C
The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released
Q in = Q out
Heat can be calculated using the formula:
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
[tex]\rm \Delta T=\dfrac{Q}{m.c}[/tex]
Because temperature change is inversely proportional to the value of c (heat capacity) and because the mass of an element / compound is equal = 10 g,and the heat absorbed also the same, then the greatest temperature change upon absorbing 100.0J of heat is the one with the smallest heat capacity (c), namely: Pb with C Pb = 0.128J / g ° C
[tex]\rm \Delta T=\dfrac{100}{10\times 0.128}\\\\\Delta T=78.125\:C[/tex]
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