Question

when 40.0 ml of 1.00 m h₂so₄ is added to 80.0 ml of 1.00 m naoh at 20.00°c in a coffee cup calorimeter, the temperature of the aqueous solution increases to 29.20°c. if the mass of the solution is 120.0 g and the specific heat of the calorimeter and solution is 4.184 j/g·°c, how much heat is given off in the reaction? (ignore the mass of the calorimeter in the calculation.) use q = mcₚδt. 4.62 kj 10.0 kj 14.7 kj 38.5 kj

Explanation:

Step1: Identify given values

We have \( m = 120.0\ g \), \( C_p = 4.184\ J/g\cdot^\circ C \), initial temperature \( T_i = 20.00^\circ C \), final temperature \( T_f = 29.20^\circ C \).

Step2: Calculate temperature change (\(\Delta T\))

\(\Delta T = T_f - T_i = 29.20^\circ C - 20.00^\circ C = 9.20^\circ C\)

Step3: Apply the formula \( q = mC_p\Delta T \)

Substitute the values: \( q = 120.0\ g \times 4.184\ J/g\cdot^\circ C \times 9.20^\circ C \)
First, calculate \( 120.0 \times 4.184 = 502.08 \)
Then, \( 502.08 \times 9.20 = 4619.136\ J \)
Convert to kJ: \( 4619.136\ J = 4.619136\ kJ \approx 4.62\ kJ \)

Answer:

4.62 kJ