# Specific heat In this problem answers are requested to three significant digits for grading...

## Question:

Specific heat In this problem answers are requested to three significant digits for grading purposes. The true number of significant digits may be more or less.

Part A

A volume of 75.0 mL of H_2O is initially at room temperature (22.00 C). A chilled steel rod at 2.00 C is placed in the water. If the final temperature of the system is 21.40 C, what is the mass of the steel bar?

Use the following values:

specific heat of water = 4.18 J/(g C)

specific heat of steel = 0.452 J/(g C)

Part B

The specific heat of water is 4.18 J/(g C). Calculate the molar heat capacity of water.

## Heat Transfer

When two objects reach thermal equilibrium one object loses energy and lowers it's temperature while the other gains energy and raises it's temperature. If no heat is lost to the surroundings this means that the heat lost by one object is gained by the other. The amount of heat that is transferred is dependant upon the mass, specific heat and the starting and final temperatures of both objects.

## Answer and Explanation:

The heat lost be the water is gained by the steel rod. This is expressed by {eq}q_w = - q_s {/eq}. Since {eq}q = mC\triangle T {/eq} the mass of the steel rod can be determined.

{eq}m_wC_w\triangle T_w = -m_sC_s\triangle T_s {/eq}

{eq}(75g)(4.18)(21.40 - 22.00 ^oC) = m_s(0.452)(2.00 - 21.40 ^oC) {/eq} note that in this step the negative sign on the right hand side was removed and was replaced by {eq}T_i - T_f {/eq}to make the calculations simpler

{eq}(75g)(4.18)(21.40 - 22.00 ^oC) / (0.452)(2.00 - 21.40 ^oC) = m_s {/eq}

{eq}m_s = 21.5 g {/eq}

B

The grams in the units of specific heat needs to be converted to moles. This can be done by dividing the specific heat by the molar mass. {eq}4.18 J/g ^oC * 18 g/mol = 75.2 J/mol ^oC {/eq}