A sample of water with a mass of 648.00 kg at 298 K is heated with 87 kJ of energy. The specific heat of water is [tex]$1 J-1 kg K -1$[/tex]. What is the final temperature of the water?
A. 298.00 K
B. 298.13 K
C. 432.26 K
D. 134.26 K
Answer (1)
Use the formula Q = m c Δ T to relate heat energy, mass, specific heat, and temperature change.
Calculate the temperature change: Δ T = m c Q = 648 × 1 87000 = 134.26 K.
Determine the final temperature: T f = T i + Δ T = 298 + 134.26 = 432.26 K.
The final temperature of the water is 432.26 K .
Explanation
Problem Setup We are given a sample of water with a mass of 648.00 kg at an initial temperature of 298 K. We add 87 kJ of energy to the water and want to find the final temperature. The specific heat of water is given as 1 J k g − 1 K − 1 .
Relevant Formula We will use the formula Q = m c Δ T , where:
Q is the heat energy added,
m is the mass of the water,
c is the specific heat of water, and
Δ T is the change in temperature ( T f − T i ).
Given Values We are given:
m = 648.00 kg
T i = 298 K
Q = 87 kJ = 87000 J
c = 1 J k g − 1 K − 1
Solving for Temperature Change We want to find T f . First, we solve for Δ T :
Δ T = m c Q Δ T = ( 648.00 k g ) ( 1 J k g − 1 K − 1 ) 87000 J
Calculating Temperature Change Δ T = 134.25925925925927 K Rounding to two decimal places, we get Δ T = 134.26 K.
Calculating Final Temperature Now we can find the final temperature T f :
T f = T i + Δ T T f = 298 K + 134.26 K T f = 432.26 K
Final Answer The final temperature of the water is 432.26 K.
Examples
Imagine you're heating water in a large tank for an industrial process. Knowing the specific heat of water allows you to calculate how much energy you need to raise the water to a desired temperature. This calculation is crucial for controlling the process and ensuring it runs efficiently. For example, if you need to heat 648 kg of water from 298 K to 432.26 K, you can determine the exact amount of energy required, which helps in designing the heating system and managing energy consumption.
