Answer to Question #223152 in Mechanics | Relativity for Consolar

Question

A piece of Copper of mass 100g is heated to 100⁰C and transferred to a well lagged Copper can of 50g containing 200g of water at 10⁰C. The final temperature of the water is 14⁰C after stirring neglecting heat losses, calculate the specific heat capacity of copper using either mixing method or electrical method. State clearly the precautions you will take. The specific heat capacity of water = 4.2 x 10²Jkg^-1K^-1

Answer 1

Explanations & Calculations

• The heat released by the piece of copper equals the heat gained by both water and the pot of copper.

$\qquad\qquad \begin{aligned} \small \frac{100}{1000}\times C_c \times (100-14)&=\small \frac{200}{1000}\times4200 \times(14-10)+\frac{50}{1000}\times C_ c\times(14-10)\\ \small 8.6C_c&=\small 3360+0.2C_c\\ \small C_c &=\small 400\,Jkg^{-1}K^{-1} \end{aligned}$

• Mass in Grams should be converted into Kilograms otherwise, the numerical values receive to be faulty.