Question

A glass piece weighing 50 grams, possessing a specific heat of 0.84 J/g°C, undergoes heating from 35°C to 46°C. What is the total amount of heat added to the glass?(1 point)
Responses

11 J
11 J

193 J
193 J

462 J
462 J

1470 J

Answer 1

To calculate the total amount of heat added to the glass, we can use the formula:

\[ Q = mc\Delta T \]

where:
- \( Q \) is the heat added (in joules),
- \( m \) is the mass (in grams),
- \( c \) is the specific heat (in J/g°C),
- \( \Delta T \) is the change in temperature (in °C).

Given:
- \( m = 50 \) grams,
- \( c = 0.84 \) J/g°C,
- Initial temperature, \( T_1 = 35 \)°C,
- Final temperature, \( T_2 = 46 \)°C.

First, we need to calculate \( \Delta T \):

\[ \Delta T = T_2 - T_1 = 46°C - 35°C = 11°C \]

Now we can substitute the values into the formula:

\[ Q = mc\Delta T \]
\[ Q = 50 \, \text{g} \times 0.84 \, \text{J/g°C} \times 11 \, \text{°C} \]
\[ Q = 50 \times 0.84 \times 11 \]
\[ Q = 50 \times 9.24 \]
\[ Q = 462 \, \text{J} \]

Therefore, the total amount of heat added to the glass is **462 J**.