Question

1. a 40 g sample of water absorbs 500 joules of energy. how much did the water temperature change? the specific heat of water (liquid) is 4.18 j/(g×°c). round to the nearest whole number

Explanation:

Step1: Recall the heat - transfer formula

The formula for heat transfer is $Q = mc\Delta T$, where $Q$ is the heat energy 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 $\Delta T$.

Step2: Rearrange the formula for $\Delta T$

From $Q = mc\Delta T$, we can get $\Delta T=\frac{Q}{mc}$.

Step3: Substitute the given values

We know that $Q = 500\ J$, $m = 40\ g$, and $c = 4.18\ J/(g\cdot^{\circ}C)$. Substituting these values into the formula $\Delta T=\frac{Q}{mc}$, we have $\Delta T=\frac{500}{40\times4.18}$.

Step4: Calculate the value of $\Delta T$

First, calculate $40\times4.18 = 167.2$. Then, $\Delta T=\frac{500}{167.2}\approx 3^{\circ}C$.

Answer:

$3^{\circ}C$