Question

per mole undergoes a temperature change of Salve AT = 98 °C - 22°C = 76°C - 10 9 = C, XmX AT = (4.18J/g-K)(250 g)(76 K) = 7.9 X 10^J 1 mol H2O = 18.0 g H20 (a) The water under in Equation 5.22, we have heat capacity is the heat capacity of one mole of sub- e atomic weights of hydrogen and oxygen, we have (180g) = 75.2 1/mol-K m = (4.18 m = (4.18 g-kli moll b) The molar heat ca stance. Using the aton pecific heat given in part (a), we have From the specific heater PRACTICE EXERCISE (a) Large beds of rock the rocks is 0.82/ creases by 12.0°C. ads of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of s 0.82 l/g-K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature in- by 12.0°C. (b) What temperature change would these rocks undergo if they emitted 450 kJ of heat? (a) 4.9 X 10 J, (b) 11 K decrease = 11 °C decrease Constant-Pressure Calorimetry GO FIGU The techniques and equipment employed in calorimetry depend on the nature of the Propose a reas process being studied. For many reactions, such as those occurring in solution, it is easy Styrofoam cu o control pressure so that AH is measured directly. Although the calorimeters used for instead of just wy accurate work are precision instruments, a simple "coffee-cup calorimeter URE 5.18) is often used in general chemistry laboratories to illustrate the princi- orimetry. Because the calorimeter is not sealed, the reaction occurs under the of the atmosphere. le containing a reactant, to a coffee-cup In this case there is FIGURE 5.18) is 01 ples of calorimetry. Because ind
I just need help with the practice excersise.

a) Large beds of rocks are used in some solar – heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/g-K. calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0 degrees Celsius. (b) what temperature change would these rocks undergo if they emitted 450 kJ of heat?

Answer 1

Answer -

Given,

Specific Heat of rocks = 0.82 J/g.K

Mass of rocks = 50.0 kg or 50,000 g

°C + 273.15 = K

a) Heat absorbed when Temperature Increase by 12 ^$\degree$ C = ?

Here the trick is temperature increase in C or K is same. So, temperature increase is 12 K

We know that,

Heat = m * s * $\Delta$ T

Put the values,

Heat absorbed = 50,000 g * 0.82 J/g.K * 12 K

Heat absorbed = 4.92 * 10⁵ J [ANSWER]

b) Temperature change when 450 kJ or 450000 J is emitted = ?

Heat = m * s * $\Delta$ T

Put the values,

450000 J = 50,000 g * 0.82 J/g.K * $\Delta$ T

$\Delta$T = (450000 J)/(50,000 g * 0.82 J/g.K)

$\Delta$T = 10.9756097561 K or 11 K approx

As the Heat is emitted so Temperature is decreased.

11 K decrease or 11 °C decrease both are same [ANSWER]