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

Question

DaniilM [7]

3 years ago

10

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?

Chemistry

2 answers:

Flauer [41] · Answer 1 · 2022-02-22T12:59:58+03:00

Heat energy is calculated by multiplying the mass, specific heat capacity of a substance by the change in temperatures. Therefore,the heat lost by water will be given by mass of water (in kg) × specific heat capacity of water × change in temperature. This heat will be equivalent to the heat gained by the spoon calculated by mass of the spoon by specific heat capacity by change in temperature. (considering that the specific heat capacity of gold is 125.6 J/kg/k)
hence, 0.05 kg × 49 × 125.6 J/kg/k = 307.72 Joules
therefore, heat lost by water is equivalent to 307.72 Joules

zimovet [89] · Answer 2 · 2022-02-22T14:31:43+03:00

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.