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 grams?
give two answers
select 2 correct answer(s)
1 g °c 1
(--------------)(--------------)
0.0305 cal 50.0 g
0.0305 cal 50.0 g
(--------------)(--------------)
1 g °c 1
(1 g °c/0.0305 cal) (1/50.0 g)
(0.0305 cal/1 g °c) (50.0 g)

Explanation:

Step1: Recall heat - formula

The formula for heat is $Q = mc\Delta T$, where $Q$ is heat, $m$ is mass, $c$ is specific heat, and $\Delta T$ is 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: Calculate heat

$Q=50.0\ g\times0.0305\ cal/g^{\circ}C\times4.0^{\circ}C = 6.1\ cal$. To cancel out grams in the unit - conversion, we use the specific - heat value which has grams in the denominator of its unit. The correct ways to cancel out grams when using the specific - heat formula are when we multiply the mass ($m$) with the specific heat ($c$) such that the units work out. The specific heat $c = 0.0305\ cal/g^{\circ}C$ and mass $m = 50.0\ g$. When we multiply $(0.0305\ cal/1\ g^{\circ}C)(50.0\ g)$, the grams unit cancels out. Also, if we consider the inverse operations for unit - conversion, we can think of it in terms of $(1\ g^{\circ}C/0.0305\ cal)(1/50.0\ g)$ which is also a valid way to analyze the unit - cancellation process in a more general unit - conversion context.

Answer:

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