College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 14 - Problems - Page 534: 23


$1.34~kg$ of water was in the vessel.

Work Step by Step

The increase in internal energy is equal to the initial gravitational potential energy of the water at a height of 100 meters: $\Delta E = mgh$ $\Delta E = (1.00~kg)(9.80~m/s^2)(100~m)$ $\Delta E = 980~J$ The internal energy increases by $980~J$ We can find the mass $m$ of water that was in the vessel: $Q = (m+1.00~kg)~c~\Delta T$ $(m+1.00~kg) = \frac{Q}{c~\Delta T}$ $m = \frac{Q}{c~\Delta T} - 1.00~kg$ $m = \frac{980~J}{(4186~J/kg~C^{\circ})(0.100~C^{\circ})} - 1.00~kg$ $m = 1.34~kg$ $1.34~kg$ of water was in the vessel.
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