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

Answer:

Mass of lead, m = 687 g

Explanation:

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

$\Delta H=m\times C\times \Delta T$

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,  

$47.0=m\times 0.0380\times 1.8$

Mass of lead, m = 687 g