Question

how much heat is required to heat 50.0 g of gold from 25.0 to 29.0 °c if the specific heat is 0.0305 cal/g°c? how do you cancel out °c? give two correct answers select 2 correct answer(s) 1 g °c 1 1 (----------------)(----------------)(----------------) 0.0305 cal 50.0 g 4.0 °c 0.0305 cal 50.0 g 4.0 °c (----------------)(----------------)(----------------) 1 g °c 1 1 (0.0305 cal/1 g °c) (50.0 g) (4.0 °c) (1 g °c/0.0305 cal) (1/50.0 g) (1/4.0 °c)

Explanation:

Step1: Identify the heat - formula

The formula for heat $Q = mc\Delta T$, where $m$ is mass, $c$ is specific heat, and $\Delta T$ is the change in temperature. Here, $m = 50.0\ g$, $c=0.0305\ cal/g^{\circ}C$, and $\Delta T=(29.0 - 25.0)^{\circ}C = 4.0^{\circ}C$.

Step2: Substitute values into the formula

$Q=(0.0305\ cal/g^{\circ}C)\times(50.0\ g)\times(4.0^{\circ}C)$. The $g$ and $^{\circ}C$ units cancel out as required.

Answer:

$(0.0305\ cal/1\ g\ ^{\circ}C)\ (50.0\ g)\ (4.0\ ^{\circ}C)$