Question

1. 100.0 g of 4.0°c water is heated until its temperature is 37°c. calculate the amount of heat energy needed to cause this rise in temperature.

Explanation:

Step1: Identify the 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 capacity, and $\Delta T$ is the change in temperature. For water, the specific - heat capacity $c = 4.18\ J/(g\cdot^{\circ}C)$.

Step2: Calculate the change in temperature

$\Delta T=T_{final}-T_{initial}=37^{\circ}C - 4^{\circ}C=33^{\circ}C$.

Step3: Substitute the values into the formula

$m = 100.0\ g$, $c = 4.18\ J/(g\cdot^{\circ}C)$, and $\Delta T = 33^{\circ}C$. Then $Q=mc\Delta T=(100.0\ g)\times(4.18\ J/(g\cdot^{\circ}C))\times33^{\circ}C$.
$Q = 100\times4.18\times33\ J=13794\ J$.

Answer:

$13794\ J$