A 50.0 g gold spoon at 10.0 °c is placed in a cup of how water 95.0 °c. how much heat does the water lose to the spoon if the sp
oon reaches a temperature of 59.0 °c?
2 answers:
The water lost 313.6 J of heat to the spoon
<h3>Further explanation </h3>Temperature is one of the principal quantities that shows the degree of heat or cold from an object / space
Temperature is used as a guide to the level of heat energy from objects
Temperature cannot be measured by the surface of the hand because it is not accurate but can be measured by a thermometer. Temperature units include Celsius, Fahrenheit, Reamur or Kelvin
Whereas heat is a form of energy that can flow from high-temperature objects to low-temperature objects.
So heat moves when there is a difference in temperature and can increase or decrease the temperature. An object that receives a lot of heat will cause a large increase in temperature
The amount of heat is influenced by the mass of the object and the difference in temperature
Can be formulated
Q = m.c.ΔT
Q: heat received or removed by an object (J)
m: object mass (kg)
c: heat type substance (J / kg⁰C)
ΔT: temperature change (⁰C)
When A 50.0 g of gold spoon at 10.0 ° c is placed in a cup of hot water 95.0 ° c. Heat balance occurs:
q lost = q gain
q lost from water = q gain from spoon
q lost from water = m. cp (gold). delta t
The specific heat of gold is 0.128 J / g ° C.
q lost from water = 50 g. 0.128. (59-10)
q lost from water = 313.6 J
<h3>Learn more </h3>
heat and temperature
grams of water
decrease heat transfer between two objects
Keywords: heat gain, heat lost
You might be interested in
Answer:
Explanation:
Hola,
En este caso, podemos usar la ley de Boyle, la cual nos permite analizar el comportamiento volumen-presión en un gas ideal de manera inversamente propocional:
Así, dado el volumen y la presión inicial, la cual se convierte a atmósferas (760 torr = 1atm), calculamos el volumen final a 1 atm como se muestra a continuación:
Saludos!