Question

A certain substance has a mass per mole of 50.0 g/mol. When 314 J is added as heat to a 30.0 g sample, the sample’s temperature rises from 25.0°C to 45.0°C. (a) What is the specific heat and (b) What is the molar specific heat of this substance? (c) How many moles are in the sample?

Short answer

1. The specific heat of the substance is$523\text{\hspace{0.17em}J/kg.K}$
2. The molar specific heat of this substance is$26.2\text{\hspace{0.17em}J/mol.K}$
3. There are$0.600$ moles in the sample.

Step-by-step solution

The given data

1. The mass per mole of the substance is$\mathrm{M}=50.0\frac{\text{g}}{\text{mol}}\text{or}50.0\times {10}^{-3}\frac{\text{kg}}{\text{mol}}$
2. The mass of the sample is$\mathrm{m}=30.0\text{\hspace{0.17em}gor}30.0\times {10}^{-3}\text{kg}$

1. The heat added to the sample is$\mathrm{Q}=314\text{\hspace{0.17em}J}$
2. The initial temperature of the substance is${\mathrm{T}}_{\mathrm{i}}=25.0°\mathrm{C}$
3. The final temperature of the substance is$T_f=45.0°C$

Understanding the concept of specific heat

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, $\mathrm{Q}=\mathrm{mc\Delta T}$ …(i)

Where, m = mass

$\mathrm{c}$= specific heat capacity

$\Delta T$= change in temperature

$Q$= required heat energy

The heat energy required by a body,$\mathrm{Q}={\mathrm{Nc}}_{\mathrm{m}}\mathrm{\Delta T}$ …(ii)

Where, N = m/M = number of moles

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

$\mathrm{\Delta T}$= change in temperature

Q= required heat energy

(a) Calculation of the specific heat of the substance

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

$\begin{array}{c}\mathrm{c}=\frac{\mathrm{Q}}{\mathrm{m}({\mathrm{T}}_{\mathrm{f}}-{\mathrm{T}}_{\mathrm{i}})}\\ =\frac{314\text{J}}{30.0\times {10}^{-3}\text{kg}\cdot (45.0\text{°C}-25.0\text{°C})}\\ =523\text{J/kg.K}\end{array}$

Hence, the value of the specific heat is$523\text{\hspace{0.17em}J/kg.K}$

(b) Calculation of the molar specific heat of the substance

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

$\begin{array}{c}{\mathrm{c}}_{\mathrm{m}}=\frac{\mathrm{Q}}{\mathrm{N}({\mathrm{T}}_{\mathrm{f}}-{\mathrm{T}}_{\mathrm{i}})}\\ =\frac{314\text{J}}{\left(\frac{30.0\times {10}^{-3}\text{kg}}{50.0\times {10}^{-3}\text{kg/mol}}\right)(45.0\text{°C}-25.0\text{°C})}\\ =26.2\text{\hspace{0.17em}J/mol.K}\end{array}$

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

(c) Calculation of the number of moles of the sample

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

$\begin{array}{c}N=\frac{m}{M}\\ =\frac{30.0\times {10}^{-3}\text{kg}}{50.0\times {10}^{-3}\text{kg/mol}}\\ =0.600\text{\hspace{0.17em}mol}\end{array}$

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