Question

1. what mass of water will change its temperature by 3°c when 525 j of heat is added to it?

Explanation:

Step1: Recall the heat formula

The formula for heat transfer 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 \( c = 4.18 \, \text{J/g}^\circ\text{C} \) (or sometimes approximated as \( 4.2 \, \text{J/g}^\circ\text{C} \) for simplicity). We need to solve for \( m \), so rearrange the formula: \( m=\frac{Q}{c\Delta T} \).

Step2: Substitute the given values

We know \( Q = 525 \, \text{J} \), \( \Delta T = 3^\circ\text{C} \), and using \( c = 4.2 \, \text{J/g}^\circ\text{C} \) (approximate value for water). Plugging into the formula: \( m=\frac{525}{4.2\times3} \).

Step3: Calculate the denominator

First, calculate \( 4.2\times3 = 12.6 \).

Step4: Calculate the mass

Then, \( m=\frac{525}{12.6}=41.67 \, \text{g} \) (or if using \( c = 4.18 \, \text{J/g}^\circ\text{C} \), \( m=\frac{525}{4.18\times3}=\frac{525}{12.54}\approx41.87 \, \text{g} \), but the approximate value with \( c = 4.2 \) is common in basic problems).

Answer:

The mass of water is approximately \( \boldsymbol{41.7 \, \text{g}} \) (or \( 41.67 \, \text{g} \) or \( 41.87 \, \text{g} \) depending on the specific heat capacity used).