Question

1. how much heat must be absorbed by a 15.0 g sample of water to raise its temperature from 25.0°c to 55.0°c? (the specific heat of water is 4.184 j/(g°c))

Explanation:

Step1: Calculate the temperature change

The formula for temperature change is $\Delta T=T_{final}-T_{initial}$.
Given $T_{initial} = 25.0^{\circ}C$ and $T_{final}=55.0^{\circ}C$, so $\Delta T=55.0 - 25.0=30.0^{\circ}C$.

Step2: Use the heat - absorption formula

The formula for heat absorption is $Q = mc\Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the temperature change.
Given $m = 15.0g$, $c = 4.184\frac{J}{g^{\circ}C}$, and $\Delta T = 30.0^{\circ}C$.
Substitute the values into the formula: $Q=(15.0g)\times(4.184\frac{J}{g^{\circ}C})\times(30.0^{\circ}C)$.
First, calculate $15.0\times4.184 = 62.76$.
Then, $62.76\times30.0=1882.8J$.

Answer:

$1882.8J$