Question

6.28 105 j of heat is supplied to raise the temperature of water from 25.0 °c to 100 °c. the specific heat of water is 4.20 j/g°c. the mass of water used is 1.55 kg 1.95 kg 1.99 kg 6.59 kg 1.50 kg

Explanation:

Step1: Recall the heat formula

The formula for heat \( Q \) is \( Q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat, and \( \Delta T \) is temperature change.

Step2: Calculate temperature change

\( \Delta T = 100^\circ C - 25.0^\circ C = 75.0^\circ C \)

Step3: Rearrange formula for mass

\( m = \frac{Q}{c\Delta T} \)

Step4: Substitute values

Given \( Q = 6.28\times10^{5}\ J \), \( c = 4.20\ J/g^\circ C \), \( \Delta T = 75.0^\circ C \).
First, convert units: \( c = 4.20\ J/g^\circ C = 4200\ J/kg^\circ C \) (since \( 1\ kg = 1000\ g \)).
Then, \( m = \frac{6.28\times10^{5}\ J}{4200\ J/kg^\circ C \times 75.0^\circ C} \)
Calculate denominator: \( 4200 \times 75.0 = 315000 \)
\( m = \frac{628000}{315000} \approx 1.99\ kg \)

Answer:

1.99 kg