91Ó°ÊÓ

1. The mass per mole of the substance is$\mathrm{M}=50.0\text{ }\frac{\text{g}}{\text{mol}}\text{or}50.0×{10}^{−3}\frac{\text{kg}}{\text{mol}}$
2. The mass of the sample is$\mathrm{m}=30.0\text{ g´Ç°ù}30.0×{10}^{−3}\text{kg}$

1. The heat added to the sample is$\mathrm{Q}=314\text{ J}$
2. The initial temperature of the substance is${\mathrm{T}}_{\mathrm{i}}=25.0°\mathrm{C}$
3. The final temperature of the substance is$T_f=45.0°C$

A substance's specific heat capacity is the amount of energy required to raise its temperature by one degree Celsius. We can use the concept of specific heat and molar specific heat of the water. We can use the expression of the number of moles of the substance. It is the ratio of the mass of the sample to the mass per mole of the substance.

Formulae:

The heat energy required by a body, $\mathrm{Q}=\mathrm{³¾³¦Î”°Õ}$ …(¾±)

Where, m = mass

$\mathrm{c}$= specific heat capacity

$\mathrm{Δ}T$= change in temperature

$Q$= required heat energy

The heat energy required by a body,$\mathrm{Q}={\mathrm{Nc}}_{\mathrm{m}}\mathrm{Δ°Õ}$ …(¾±¾±)

Where, N = m/M = number of moles

${\mathrm{c}}_{\mathrm{m}}$= molar specific heat capacity

$\mathrm{Δ°Õ}$= change in temperature

Q= required heat energy

Using equation (i), the specific heat of a substance is given as:

$\begin{array}{c}\mathrm{c}=\frac{\mathrm{Q}}{\mathrm{m}({\mathrm{T}}_{\mathrm{f}}−{\mathrm{T}}_{\mathrm{i}})}\\ =\frac{314\text{J}}{30.0×{10}^{−3}\text{kg}â‹\dots (45.0\text{°°ä}−25.0\text{°°ä})}\\ =523\text{ }\text{J/kg.K}\end{array}$

Hence, the value of the specific heat is$523\text{ J/°ì²µ.°­}$

The expression of the molar specific heat of the substance using equation (ii) is given as:

$\begin{array}{c}{\mathrm{c}}_{\mathrm{m}}=\frac{\mathrm{Q}}{\mathrm{N}({\mathrm{T}}_{\mathrm{f}}−{\mathrm{T}}_{\mathrm{i}})}\\ =\frac{314\text{J}}{\left(\frac{30.0×{10}^{−3}\text{kg}}{50.0×{10}^{−3}\text{kg/mol}}\right)(45.0\text{°°ä}−25.0\text{°°ä})}\\ =26.2\text{ J/³¾´Ç±ô.°­}\end{array}$

Hence, the value of the molar specific heat is$26.2\text{ }\text{J/mol.K}$

The number of moles of the sample is$N$. Hence, using equation (ii), it is given as:

$\begin{array}{c}N=\frac{m}{M}\\ =\frac{30.0×{10}^{−3}\text{kg}}{50.0×{10}^{−3}\text{kg/mol}}\\ =0.600\text{″¾´Ç±ô}\end{array}$

Hence, there are $0.600\text{mol}$in the substance