Question

1. calculate the specific heat capacity of a piece of wood if 1500.0 g of the wood absorbs 67,500 joules of heat, and its temperature changes from 32°c to 57°c.

Explanation:

Step1: Identify the heat - transfer formula

The heat - transfer formula 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 $c$.

Step2: Calculate the change in temperature

$\Delta T=T_f - T_i$, where $T_f = 57^{\circ}C$ and $T_i = 32^{\circ}C$. So, $\Delta T=57 - 32=25^{\circ}C$.

Step3: Rearrange the formula to solve for $c$

From $Q = mc\Delta T$, we can solve for $c$ as $c=\frac{Q}{m\Delta T}$. Given $Q = 67500\ J$, $m = 1500.0\ g$, and $\Delta T = 25^{\circ}C$.
Substitute the values into the formula: $c=\frac{67500\ J}{1500.0\ g\times25^{\circ}C}$.

Step4: Perform the calculation

$c=\frac{67500}{1500.0\times25}=\frac{67500}{37500}=1.8\ J/(g\cdot^{\circ}C)$

Answer:

$1.8\ J/(g\cdot^{\circ}C)$