Question

1. what is the heat gained by an insulated beaker of water when a cube of copper is dropped into the water. the mass of the water is 375 grams. the initial temperature of the water is 23 °c and the final temperature of the water is 52 °c. the specific heat of the water is 4184 j/kg-°c

Explanation:

Step1: Convert mass to kg

$m = 375\ \text{g} = 0.375\ \text{kg}$

Step2: Calculate temperature change

$\Delta T = 52^\circ\text{C} - 23^\circ\text{C} = 29^\circ\text{C}$

Step3: Apply heat gain formula

$Q = mc\Delta T$
$Q = 0.375\ \text{kg} \times 4184\ \text{J/kg-}^\circ\text{C} \times 29^\circ\text{C}$

Step4: Compute the final value

$Q = 0.375 \times 4184 \times 29 = 45729\ \text{J}$

Answer:

$45729\ \text{J}$