Question

The specific heat of lead is 0.0380 cal/g·°C. If 47.0 calories of energy raised the temperature of a lead sample from 28.3°C to 30.1°C what is the mass of the sample?

Answer 1

Mass of lead, m = 687 g

Step-by-step explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

`\Delta H=m* C* \Delta T`

Where,

`\Delta H` is the enthalpy change

m is the mass

C is the specific heat capacity

`\Delta T` is the temperature change

Thus, given that:-

Mass = ?

Specific heat = 0.0380 cal/g°C

`\Delta T=30.1-28.3\ ^0C=1.8\ ^0C`

Heat added = 47.0 calories

So,

`47.0=m* 0.0380* 1.8`

Mass of lead, m = 687 g