Question

Calculate the heat transferred when 50 g of water is heated from 20\(^\circ\)C to 80\(^\circ\)C. The specific heat capacity of water is (4)18 J/\(^\circ\)C\(^{-1}\)g\(^{-1}\)

• A. 12,540 J
• B. 10,450 J
• C. 8,360 J
• D. 6,270 J

Hint:

Specific Heat Calculation. Heat required to change temperature: \(Q = mc\Delta T\). Ensure units are consistent (mass in g if c is per g, or mass in kg if c is per kg). \(\Delta T\) can be in K or °C as it's a difference.

Answer 1

The Correct Option is A

The heat transferred (\(Q\)) when changing the temperature of a substance is given by the formula: $$ Q = mc\Delta T $$ where \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature (\(T_{final} - T_{initial}\)).
Given: Mass \(m = 50\) g Specific heat capacity \(c = (4)18\) J/\(^\circ\)C/g Initial temperature \(T_i = 20^\circ\)C Final temperature \(T_f = 80^\circ\)C Change in temperature \(\Delta T = T_f - T_i = 80^\circ\text{C} - 20^\circ\text{C} = 60^\circ\)C.
Substitute the values: $$ Q = (50 \, \text{g}) \times ((4)18 \, \text{J/}^\circ\text{C/g}) \times (60 \, ^\circ\text{C}) $$ $$ Q = 50 \times (4)18 \times 60 \, \text{J} $$ $$ Q = 50 \times 250.
8 \, \text{J} $$ $$ Q = 12540 \, \text{J} $$ The heat transferred is 12,540 J.