Answer:
1.2 × 10⁴ cal
Explanation:
Given data
- Initial temperature: 80 °C
We can calculate the heat released by the water (
) when it cools using the following expression.

where
c is the specific heat capacity of water (1 cal/g.°C)

According to the law of conservation of energy, the sum of the heat released by the water (
) and the heat absorbed by the reaction (
) is zero.

Use Planck's equation (E=hv) to solve. where <span>frequency (v) of ultrviolet radiation is 6.8 × 1015 1/s. </span><span>
</span>The variable h is a
constant equal to 6.63 × 10-34 J·s
E= <span>(6.8 × 1015 1/s)x(</span>6.63 × 10-34 J·s)
Convection is where the heat source heats the air, and the air transfers the heat to object. Some people prefer to heat their marshmallow through convection, this is achieved by holding the marshmallow high above the flames where the heated air is rising, transforing the heat into the marshmallow.