Question

question 23 of 30 if 11 kj of heat is transferred to 2300 g of soil at 21°c, what would the new temperature of the soil be? (the specific heat capacity of soil is 0.80 j/g·°c.) a. 21°c b. 6°c c. 27°c d. 109°c

Explanation:

Step1: Convert heat to joules

The heat transferred \( Q = 11\space kJ = 11\times1000\space J = 11000\space J \)

Step2: Use heat formula \( Q = mc\Delta T \) to find \( \Delta T \)

We know \( Q = 11000\space J \), \( m = 2300\space g \), \( c = 0.80\space J/g\cdot^\circ C \)
Rearranging the formula for \( \Delta T \): \( \Delta T=\frac{Q}{mc} \)
Substitute the values: \( \Delta T=\frac{11000}{2300\times0.80} \)
Calculate denominator: \( 2300\times0.80 = 1840 \)
Then \( \Delta T=\frac{11000}{1840}\approx 6^\circ C \)

Step3: Find new temperature

Initial temperature \( T_i = 21^\circ C \), new temperature \( T_f=T_i+\Delta T \)
\( T_f = 21 + 6 = 27^\circ C \)

Answer:

C. \( 27^\circ C \)