Question

400.0 g of a metal absorbs 10000. j of heat energy and its temperature rises from 20.0 °c to 103.0 °c. what is the specific heat of the metal?
0.301 j/g°c
0.255 j/g°c
3.32 j/g°c
0.243 j/g°c
question 10
6 pts
250.0 g of a metal releases 5000. j of energy and its temperature drops from 90.0 °c to 15.7 °c. what is the specific heat of the metal?
3.72 j/g°c
0.269 j/g°c
0.222 j/g°c
1.27 j/g°c

Explanation:

Step1: Recall heat - capacity formula

The formula for heat energy is $Q = mc\Delta T$, where $Q$ is the heat energy, $m$ is the mass, $c$ is the specific heat, and $\Delta T$ is the change in temperature. We can re - arrange it to solve for $c$: $c=\frac{Q}{m\Delta T}$.

Step2: Calculate $\Delta T$ for the first question

For the first problem, $\Delta T=T_2 - T_1=103.0^{\circ}C - 20.0^{\circ}C = 83.0^{\circ}C$, $m = 400.0g$ and $Q = 10000J$. Then $c=\frac{Q}{m\Delta T}=\frac{10000J}{400.0g\times83.0^{\circ}C}\approx0.301J/g^{\circ}C$.

Step3: Calculate $\Delta T$ for the second question

For the second problem, $\Delta T=T_1 - T_2=90.0^{\circ}C - 15.7^{\circ}C = 74.3^{\circ}C$, $m = 250.0g$ and $Q = 5000J$. Then $c=\frac{Q}{m\Delta T}=\frac{5000J}{250.0g\times74.3^{\circ}C}\approx0.269J/g^{\circ}C$.

Answer:

First question: 0.301 J/g°C
Second question: 0.269 J/g°C