Basics of Heat and Temperature Changes
Problem Statement
Basics of Heat and Temperature Changes
Problem Statement
Consider the relationship between heat and temperature change when water in a cup is heated or cooled. Answer the following questions.
(1) A cup with a heat capacity of C0 contains water. How much heat, in J, is required to raise the temperature of the entire cup by ΔT0?
(2) The temperature of a mass m1 of water was raised from Ta to Tb. What is the amount of heat, in J, that the water absorbed? Let the specific heat capacity of water be cw.
(3) A constant amount of heat Q2 was applied to a mass m2 of water in a container using an electric heater, causing the temperature of the water to rise from Tc to Td. What is the specific heat of this water in J/(kg⋅K)? Assume that the heat capacity of the container and heat transfer to the outside can be neglected.
Constraints
・C0=80 J/K ・ΔT0=15 K ・m1=0.20 kg ・Ta=280 K ・Tb=295 K ・cw=4200 J/(kg\cdotK) ・m2=0.50 kg ・Q2=31500 J ・Tc=285 K ・Td=300 K
Input Format
Let the answer to Part (1) be A1 J, the answer to Part (2) be A2 J, and the answer to Part (3) be A3 J/(kg⋅K). As your final answer, calculate the value A1+A2−A3 and enter the resulting natural number as is.
Solution
Explanation
Application of Physical Laws
The amount of heat Q [J] required to change the temperature of a substance is given by the following equation, where m [kg] is the mass of the substance, c [J/(kg⋅K)] is its specific heat capacity, and ΔT [K] is the change in temperature. Q=mcΔT
Furthermore, if the heat capacity of an object is C [J/K], the amount of heat required, Q, is given by the following equation: Q=CΔT
Setting Up Equations and Calculations for Each Subquestion
Subquestion (1)
Given a heat capacity of C0=80 J/K and a temperature rise of ΔT0=15 K, the required heat A1 is calculated as follows: A1=C0×ΔT0=80×15=1200 J
Subquestion (2)
Given that the mass of the water is m1=0.20 kg, the specific heat capacity is cw=4200 J/(kg\cdotK), and the temperature change is ΔT=Tb−Ta=295−280=15 K, the amount of heat absorbed by the water, A2, is calculated as follows. A2=m1cw(Tb−Ta)=0.20×4200×15=12600 J
Subquestion (3)
Given the mass m2=0.50 kg, the heat supplied Q2=31500 J, and the temperature change ΔT′=Td−Tc=300−285=15 K, the specific heat A3 of the water is given by: A3=m2(Td−Tc)Q2=0.50×1531500=7.531500=4200 J/(kg\cdotK)
Conversion to Input Format
Substitute the calculated values into the given equation.
A1+A2−A3=1200+12600−4200 =13800−4200=9600
This results in a unique, exact integer value (natural number) without the need for rounding or approximation.
The natural number to enter is 9600.