Question

1. 45 grams of an unknown substance undergoes a temperature increase of 38 °c after absorbing 4172.4 joules. what is the specific heat of the substance? round to the nearest hundredth

Explanation:

Step1: Recall the heat - transfer formula

The formula for heat transfer is $Q = mc\Delta T$, where $Q$ is the heat absorbed or released, $m$ is the mass of the substance, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. We need to solve for $c$. Rearranging the formula gives $c=\frac{Q}{m\Delta T}$.

Step2: Identify the given values

We are given that $Q = 4172.4$ J, $m = 45$ g, and $\Delta T=38^{\circ}C$.

Step3: Substitute the values into the formula

$c=\frac{4172.4}{45\times38}$
First, calculate the denominator: $45\times38 = 1710$.
Then, calculate $c$: $c=\frac{4172.4}{1710}= 2.44$ J/(g·°C).

Answer:

$2.44$ J/(g·°C)