You burn a 15g jellybean to warm 50 mL of water, which increases the temperature of the water 25 °C. How many calories of heat are transferred from the jellybean to the water? Assume there is no heat loss.

Explanation:

The quantity of heat transfered from the jellybean to the water is:

1.25 Kcal

Explanation:

The relationship between heat and temperature change is given by the equation:

where H = heat energy (Joules, J) , m = mass of a substance (kg) , c = specific heat (units J/kg∙K) and ΔT is the change in temperature

Given that:

mass of water (m) = density of water × volume = 1g/ml × 50ml = 50 g, ΔT = 25° C and of water= 4.18J/g/°C

Therefore H = 50g × 4.18J/g/°C × 25°C = 5225 J

The heat transferred to the Jelly bean = 5225 J = 1.25 Kcal

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

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