Question

what is the specific heat capacity of a substance of mass 151.9 g that increases in temperature from 22 °c to 26.8 °c after absorbing 2.05 kj of energy in the form of heat? express your answer in j g⁻¹ °c⁻¹. do not try to identify the substance. answer:

Explanation:

Step1: Convert energy to joules

We know that $1\ kJ = 1000\ J$, so $2.05\ kJ=2.05\times1000 = 2050\ J$.

Step2: Calculate change in temperature

$\Delta T=T_2 - T_1$, where $T_1 = 22^{\circ}C$ and $T_2=26.8^{\circ}C$. So $\Delta T=26.8 - 22=4.8^{\circ}C$.

Step3: Use the specific - heat - capacity formula

The formula for heat $Q = mc\Delta T$, where $Q$ is the heat energy, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. We want to find $c$, so $c=\frac{Q}{m\Delta T}$. Substitute $Q = 2050\ J$, $m = 151.9\ g$, and $\Delta T = 4.8^{\circ}C$ into the formula: $c=\frac{2050}{151.9\times4.8}$.

Step4: Calculate the value of $c$

$151.9\times4.8 = 729.12$, and $\frac{2050}{729.12}\approx2.81\ J\ g^{-1}\ ^{\circ}C^{-1}$.

Answer:

$2.81$