I will use this formula, the convert.
q( in Joules ) = mass * specific heat Au * change in temp.
q = nCT
q = (150 grams)(0.129 J/gC)(175 C - 25 C)
= 2902.5 Joules
Now, 1 calorie = 4.184 Joules
2902.5 Joules (1 calorie/4.184 Joules)
694 calories required
When the temperature of a sample of water is -5 degrees Celsius, the water is frozen and in a solid state.
I will work in Joules, then convert to calories.q(Joules) = mass * specific heat * change in temperature Cq = (27.3 g Cu)(0.385 J/gC)(74.4o C - 36.2o C)= 402 Joules (1 calorie/4.184 Joules)= 96.1 calories===========
85 + 273 = 358 K.
The temperature of the water is 100 degrees celsius.
The change in temperature is 25 degrees Celsius, meaning it takes 22.48 joules per degree of change. The specific heat of iron is 0.449 J/g degree Celsius. This means that the mass of iron must be 50.07 grams
80
38 cal
80 calories per gram
Because they are the same temperature. There are more calories of heat in a 100-ml sample of water that's at a certain temperature than there are in a 10-ml sample that's the same temperature...but 20-degree-Celsius water is 20-degree-Celsius water whether you have a thimbleful, a glassful or a swimming pool full.
When the temperature of a sample of water is -5 degrees Celsius, the water is frozen and in a solid state.
The temperature difference in Kelvin is the same as in Celsius. So, if the sample rises by 12 degrees Celsius, it also rises by 12 Kelvin.
How much heat (in calories) is required to heat a 43 g sample of aluminum from 72 F to 145F
I will work in Joules, then convert to calories.q(Joules) = mass * specific heat * change in temperature Cq = (27.3 g Cu)(0.385 J/gC)(74.4o C - 36.2o C)= 402 Joules (1 calorie/4.184 Joules)= 96.1 calories===========
85 + 273 = 358 K.
To calculate the amount of ice water needed to cool the sample to 20 degrees Celsius, you would need the initial temperature of the sample, the mass of the sample, and the specific heat capacities of water and ice. With this information, you could use the equation q = m * c * ΔT to determine the quantity of ice water needed to cool the sample.
one calorie of heat is able to raise one gram of water one degree Celsius so 400 calories could raise 1g of water 400 degrees, so it would raise the 80g by(400/80) 5 degrees Celsius plus the initial temp of 10 degrees, the 80g of water would have a final temp of 15 degrees Celsius
The temperature of the water is 100 degrees celsius.