What is the identity of a 100. g sample of metal that, upon absorbing 4680 J of heat, increases in temperature by 52.0°C?

Aluminum is a metal that, in a 100. g sample absorbing 4680 J of heat increases in temperature by 52.0 °C.

We have a mass (m) of 100. g of an unknown metal that, when absorbs 4680 J of heat (Q) increases in temperature by 52.0°C (ΔT). The specific heat capacity (c) of the metal is the amount of heat required to raise the temperature of the unit mass by the unit of temperature. We can calculate it using the following expression.

Aluminum is the metal whose specific heat capacity is 0.900 J/g.°C.

Aluminum is a metal that, in a 100. g sample absorbing 4680 J of heat increases in temperature by 52.0 °C.

You can learn more about specific heat capacity here: link

The metal is aluminium

Explanation:

Given:

Heat, q = 4680 J

Mass, m = 100g = 0.1kg

ΔT = 52°C

sample = ?

We know:

q = mcΔT

On substituting the value we get:

Thus, the metal is aluminium which has a specific heat capacity of 900 J/kg°C

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answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol