Find the heat capacity of a pan of mass 200 g, if its temperature rises by 8° C on receiving 2000 J of heat.

Concept:

Heat capacity:

• It is the ratio of the amount of heat supplied to a change in temperature while specific heat capacity is the amount of heat supplied per unit mass per unit change in temperature.
• Formula, Heat capacity, \(C=\frac{Q}{Δ T}\), where Q = heat, ΔT = temperature difference
• Heat capacity can be measured in JK^-1.

Calculation:

Given: mass, m = 200 g = 0.2 kg, Temperature difference, ΔT = 8ºC,Heat, Q = 2000 J

Heat capacity, \(C=\frac{Q}{Δ T}\)

\(C=\frac{2000}{8}=250 JK^{-1}\)

Hence, the heat capacity is 250 JK^-1.

Mistake Points

• The difference between two temperature values in degrees Celsius and degrees Kelvin remains constant, as the additional factor of 273 in the conversion formula cancels out during subtraction.
• For example, if we have two values T₁​ and T₂​ in degrees Celsius, the difference (T1​​−T2) will be the same as the absolute difference (T1​+273−(T2​+273)) in degrees Kelvin.
• Using this example of T1​​=10∘C and T2​=18∘C, the difference in degrees Celsius is 10−18=8.
• When converted to Kelvin, the values become T1​​=283K and T2​=291K,
• With the same difference 283−291=8.

Additional Information

Specific heat capacity:

• It can be defined as the amount of heat required to change its temperature by one degree.
• Mathematically, heat required, Q = mcΔT,
• where m = mass, c = specific heat capacity, ΔT = temperature difference
• The heat can be measured in Joule (J).