A 2.00-kg block of copper at 22.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogenboil away by the time the copper reaches 77.3 K? (The specific heat of copper is 0.092 0 cal/g °C, and the latent heat ofvaporization of nitrogen is 48.0 cal/g.)kgNeed Help?Read ItWatch It

Answered: A 2.00-kg block of copper at 22.0°C is…

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter20: The First Law Of Thermodynamics

Section: Chapter Questions

Problem 20.10OQ: A 100-g piece of copper, initially at 95.0C, is dropped into 200 g of water contained in a 280-g...

See similar textbooks

Related questions

Question

A 2.00-kg block of copper at 22.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogen
boil away by the time the copper reaches 77.3 K? (The specific heat of copper is 0.092 0 cal/g °C, and the latent heat of
vaporization of nitrogen is 48.0 cal/g.)
kg
Need Help?
Read It
Watch It

Expert Solution

Step 1

When a hot body is connected to a cold body then there occurs a transfer of heat between the two objects until here occurs a thermal equilibrium. The total heat amount gained by the cold body, as per the calorimetry principle, is equal to the heat amount lost by the hot body.

Now the formula to calculate the amount of heat lost by the copper is given by

$Q_{l o s t} = m c ∆ t$

where $m$ is the copper block's mass, $c$ is the specific heat and $∆ t$ is the temperature change.

The initial temperature of the block is

$T_{i} = 22 . 0^{\circ} C$

Final temperature of the block is

$T_{f} = {(77.3 - 273.15)}^{\circ} C = - 195 . 85^{\circ} C$

As it can be seen that $T_{i} > T_{f}$ thus the temperature of the block drops.

Now the formula to calculate the change in temperature is given by

$∆ T = T_{i} - T_{f}$

Step by step

Solved in 4 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions