Question

How much heat is required to raise the temperature of $ 2 \, kg $ of water from $ 20^\circ C $ to $ 80^\circ C $? (Specific heat capacity of water = $ 4200 \, J/kg^\circ C $)

• A. \(504000 \, J\)
• B. \(50400 \, J\)
• C. \(126000 \, J\)
• D. \(168000 \, J\)

Hint:

Tip: Make sure to use consistent units and identify the temperature change correctly.

Answer 1

The Correct Option is A

To determine the amount of heat required to raise the temperature of the water, we use the formula for heat transfer: \( Q = mc\Delta T \), where:

• \( Q \) is the heat energy (in Joules),
• \( m \) is the mass of the water (in kg),
• \( c \) is the specific heat capacity (in J/kg°C),
• \( \Delta T \) is the change in temperature (in °C).

Given:

• \( m = 2 \, kg \),
• \( c = 4200 \, J/kg^\circ C \),
• Initial temperature \( T_1 = 20^\circ C \),
• Final temperature \( T_2 = 80^\circ C \).

The change in temperature is calculated as \( \Delta T = T_2 - T_1 = 80^\circ C - 20^\circ C = 60^\circ C \).

Substitute the values into the formula: \( Q = 2 \times 4200 \times 60 \).

Calculate:

• \( 2 \times 4200 = 8400 \),
• \( 8400 \times 60 = 504000 \, J \).

Therefore, the heat required to raise the temperature of 2 kg of water from \( 20^\circ C \) to \( 80^\circ C \) is \( 504000 \, J \).

Answer 2

Step 1: Write known quantities 
Mass, \( m = 2 \, kg \)
Temperature change, \( \Delta T = 80 - 20 = 60^\circ C \)
Specific heat capacity, \( c = 4200 \, J/kg^\circ C \)

Step 2: Use heat formula 
\[ Q = mc \Delta T \]

Step 3: Substitute values 
\[ Q = 2 \times 4200 \times 60 = 504000 \, J \]