A 9.02-g sample of liquid water at 40.3 °C is heated by the addition of 203.93 J of energy. The final temperature of the water is ________ °

Question

A 9.02-g sample of liquid water at 40.3 °C is heated by the addition of 203.93 J of energy. The final temperature of the water is ________ °C. The specific heat capacity of liquid water is 4.184 J/gK.

in progress 0
Adela 2 weeks 2021-08-30T01:19:51+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-08-30T01:21:16+00:00

    Answer:

    45.70 C

    Explanation:

    We’ll use the formula for specific heat Q=mcΔT. Let’s first check out units and look at the values we have.

    Q=heat= 203.93 J

    m=mass= 9.02g

    c= specific heat= 4.184 J/g K

    ΔT= Change in temperature= Final temp. – initial temp. = T_{f}T_{i}

    Let’s plug in what we know.

    203.93 J = (9.02g)(4.184)( T_{f} – 40.3 C)

    We’ll just solve this algebraically to solve for  T_{f}. It’s like solving for x, just a different variable. It’s easier to do the more complicated side first and then start moving your numbers to the other side.

    203.93 = (37.73)( T_{f} – 40.3) We’ll multiply 37.73 by  T_{f} AND 40.3 so we’ll get…

    203.93 = (37.73Tf – 1520.5)

    Let’s start moving numbers to the other side

    203.93 + 1520.5 = 37.73Tf

    1724.4 = 37.73Tf

    \frac{1724.4}{37.73}= T_{f}

    45.70 C= T_{f}

    This makes sense with our problem since they heated water, so we should see an increase in temperature.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )