Question

question 8 (1 point)
6.28 10^5 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
question 9 (1 point)

Explanation:

Step1: Calculate temperature change

$\Delta T=T_{final}-T_{initial}=100 - 25=75^{\circ}C$

Step2: Rearrange heat formula $Q = mc\Delta T$ for mass $m$

$m=\frac{Q}{c\Delta T}$

Step3: Substitute values

$Q = 6.28\times10^{5}J$, $c = 4.20J/g^{\circ}C$, $\Delta T=75^{\circ}C$
$m=\frac{6.28\times 10^{5}}{4.20\times75}$
$m=\frac{6.28\times 10^{5}}{315}\approx1993.65g$

Step4: Convert grams to kilograms

Since $1kg = 1000g$, $m=\frac{1993.65}{1000}\approx1.99kg$

Answer:

1.99 kg