Sure, let's work through this problem step by step.
Given:
- The mass of the water: [tex]\( m = 1200 \)[/tex] grams.
- The initial temperature of the water: [tex]\( T_{\text{initial}} = 36^\circ C \)[/tex].
- The amount of heat added: [tex]\( Q = 75 \, \text{kJ} = 75000 \, \text{J} \)[/tex].
- The specific heat capacity of water: [tex]\( c = 4.186 \, \text{J/g} \cdot{}^\circ C \)[/tex].
To Find:
The final temperature of the water, [tex]\( T_{\text{final}} \)[/tex].
Approach:
We will use the formula for heat transfer, which is:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
where:
- [tex]\( Q \)[/tex] is the heat added,
- [tex]\( m \)[/tex] is the mass,
- [tex]\( c \)[/tex] is the specific heat capacity,
- [tex]\( \Delta T \)[/tex] is the change in temperature ([tex]\( T_{\text{final}} - T_{\text{initial}} \)[/tex]).
First, we need to solve for [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = \frac{Q}{m \cdot c} \][/tex]
Now, let's plug in the given values:
[tex]\[ \Delta T = \frac{75000 \, \text{J}}{1200 \, \text{g} \cdot 4.186 \, \text{J/g} \cdot^\circ C} \][/tex]
Calculate the denominator first:
[tex]\[ 1200 \, \text{g} \cdot 4.186 \, \text{J/g} \cdot^\circ C = 5023.2 \, \text{J/}^\circ C \][/tex]
Next, calculate [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = \frac{75000 \, \text{J}}{5023.2 \, \text{J/}^\circ C} \approx 14.93^\circ C \][/tex]
Now, we can find [tex]\( T_{\text{final}} \)[/tex]:
[tex]\[ T_{\text{final}} = T_{\text{initial}} + \Delta T \][/tex]
[tex]\[ T_{\text{final}} = 36^\circ C + 14.93^\circ C \approx 50.93^\circ C \][/tex]
We round this to the nearest whole number, getting [tex]\( T_{\text{final}} \approx 51^\circ C \)[/tex].
Conclusion:
After adding 75 kJ of heat to 1200 grams of water initially at [tex]\( 36^\circ C \)[/tex], the new temperature of the water will be approximately [tex]\( 51^\circ C \)[/tex].
Therefore, the correct answer is:
B. [tex]\( 51^\circ C \)[/tex]
