A 2 kg object is heated from 20°C to 80°C. If the specific heat capacity is 0.5

₹0.0

Question: A 2 kg object is heated from 20°C to 80°C. If the specific heat capacity is 0.5 kJ/kg°C, how much heat is absorbed? (2023)

Options:

Correct Answer: 40 kJ

Exam Year: 2023

Solution:

Q = mcΔT = 2 kg * 0.5 kJ/kg°C * (80°C - 20°C) = 60 kJ.

A 2 kg object is heated from 20°C to 80°C. If the specific heat capacity is 0.5 kJ/kg°C, how much heat is absorbed? (2023)

A 2 kg object is heated from 20°C to 80°C. If the specific heat capacity is 0.5 kJ/kg°C, how much heat is absorbed? (2023)

Step 1: Identify the mass of the object (m). In this case, m = 2 kg.
Step 2: Identify the specific heat capacity (c). Here, c = 0.5 kJ/kg°C.
Step 3: Determine the initial temperature (T1) and final temperature (T2). T1 = 20°C and T2 = 80°C.
Step 4: Calculate the change in temperature (ΔT) using the formula ΔT = T2 - T1. So, ΔT = 80°C - 20°C = 60°C.
Step 5: Use the formula for heat absorbed (Q): Q = mcΔT.
Step 6: Substitute the values into the formula: Q = 2 kg * 0.5 kJ/kg°C * 60°C.
Step 7: Calculate the result: Q = 2 * 0.5 * 60 = 60 kJ.

Specific Heat Capacity – The amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius.
Heat Transfer Calculation – Using the formula Q = mcΔT to calculate the heat absorbed or released during a temperature change.
Unit Conversion – Understanding the units involved, particularly ensuring consistency between kJ and J if necessary.