Chapter U5, Problem 6STP

Interpretation Introduction

Interpretation:

The hottest cup of water needs to be identified, if the initial temperature of all the cups is the same.

Concept introduction:

Due to transfer of energy the temperature of any substance increases or decreases. The change in temperature for the addition of energy can be calculated with the help of calorimetric equation:

$\begin{array}{l}\text{Q = m × C × ΔT}\\ \text{Here:}\\ \text{Q}\text{}\text{ = heat change }\\ \text{m = mass}\\ \text{C = specific heat }\\ \text{ΔT = change in T}\end{array}$

Answer to Problem 6STP

Option (D) is correct statement.

Explanation of Solution

Reason for correct option:

$\begin{array}{l}\text{Q = m × C × ΔT}\\ \text{Here:}\\ \text{Q}\text{}\text{ = heat change }\\ \text{m = mass}\\ \text{C = specific heat }\\ \text{ΔT = change in T}\end{array}$

Specific heat of water = 1 calorie/gram °C 

Density of water = 1 g/mL

Thus volume of water = mass of water

Calculate the temperature for each statement:

(A)

$\begin{array}{l}\text{q = m × C × ΔT}\\ \text{160 cal = 32 g × 1 cal/gram °C × ΔT}\\ \text{ΔT=}\frac{\text{160 cal}}{\text{32 g × 1 cal/gram °C}}\\ \text{ΔT=5 °C}\end{array}$

(B)

$\begin{array}{l}\text{q = m × C × ΔT}\\ \text{525 cal = 105 g × 1 cal/gram °C × ΔT}\\ \text{ΔT=}\frac{\text{525 cal}}{\text{105 g × 1 cal/gram °C}}\\ \text{ΔT=5 °C}\end{array}$

(C)

$\begin{array}{l}\text{q = m × C × ΔT}\\ \text{2375 cal = 475 g × 1 cal/gram °C × ΔT}\\ \text{ΔT=}\frac{\text{2375 cal}}{\text{475 g × 1 cal/gram °C}}\\ \text{ΔT=5 °C}\end{array}$

Conclusion

Reasons for incorrect options:

Since the temperature difference is same for all the cups, therefore, the final temperature must be same for all the cups as the initial temperature is same for all.