Question

additional practice 1. when you turn on the hot water to wash dishes, the water pipes heat up. how much thermal energy is absorbed by a copper water pipe with a mass of 2.3 kg when its temperature is raised from 20.0°c to 80.0°c? 2. electrical power companies sell electrical energy by the kilowatt - hour, where 1 kwh = 3.6×10^6 j. suppose that it costs $0.15 per kwh to run your electric water heater. how much does it cost to heat 75 kg of water from 15°c to 43°c to fill a bathtub? 3. challenge a car engine’s cooling system contains 20.0 l of water (1 l of water has a mass of 1 kg). a. what is the change in the temperature of the water if 836.0 kj of thermal energy is added? b. suppose that it is winter, and the car’s cooling system is filled with methanol. the density of methanol is 0.80 g/cm³. what would be the increase in temperature of the methanol if it absorbed 836.0 kj of thermal energy? c. which coolant, water or methanol, would better remove thermal energy from a car’s engine? explain.

Explanation:

Problem 1

Step1: Recall the specific heat formula

The formula for thermal energy absorbed \( Q \) is \( Q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is temperature change. For copper, \( c_{copper}=387\ J/(kg\cdot^\circ C) \), \( m = 2.3\ kg \), \( \Delta T=80.0^\circ C - 20.0^\circ C = 60.0^\circ C \).

Step2: Substitute values into the formula

\( Q = 2.3\ kg\times387\ J/(kg\cdot^\circ C)\times60.0^\circ C \)
\( Q = 2.3\times387\times60\ J \)
\( Q = 2.3\times23220\ J \)
\( Q = 53406\ J = 53.4\ kJ \) (approx)

Step1: Calculate thermal energy needed

First, find \( Q \) for water: \( c_{water}=4186\ J/(kg\cdot^\circ C) \), \( m = 75\ kg \), \( \Delta T = 43^\circ C - 15^\circ C = 28^\circ C \).
\( Q = mc\Delta T = 75\ kg\times4186\ J/(kg\cdot^\circ C)\times28^\circ C \)
\( Q = 75\times4186\times28\ J \)
\( Q = 75\times117208\ J = 8790600\ J \)

Step2: Convert energy to kWh

Since \( 1\ kWh = 3.6\times10^{6}\ J \), number of kWh \( n=\frac{8790600\ J}{3.6\times10^{6}\ J/kWh}\approx2.4418\ kWh \)

Step3: Calculate cost

Cost = \( n\times0.15\ \$/\text{kWh} = 2.4418\times0.15\ \$ \approx 0.366\ \$ \) (approx \( \$0.37 \))

Step1: Find mass of water

Volume \( V = 20.0\ L = 20.0\ dm^{3}=0.02\ m^{3} \), density of water \(
ho = 1000\ kg/m^{3} \), so \( m=
ho V = 1000\ kg/m^{3}\times0.02\ m^{3}=20\ kg \).

Step2: Use \( Q = mc\Delta T \) to find \( \Delta T \)

\( c_{water}=4186\ J/(kg\cdot^\circ C) \), \( Q = 836000\ J \) (since \( 836.0\ kJ = 836000\ J \))
\( \Delta T=\frac{Q}{mc}=\frac{836000\ J}{20\ kg\times4186\ J/(kg\cdot^\circ C)} \)
\( \Delta T=\frac{836000}{83720}\approx10.0^\circ C \)

Answer:

The thermal energy absorbed is \( \boldsymbol{5.34\times10^{4}\ J} \) (or \( 53.4\ kJ \))

Problem 2